标致2008 漏油-江门 2019年医学专业招生

首页

AD联系:507867812

标致2008 漏油

时间:2019-11-20 07:08:38 作者:假日国际娱乐是骗人的吗 浏览量:97682

标致2008 漏油

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

,见下图

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇),见下图

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

,如下图

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

如下图

ajax调用返回php接口返回json数据的方法(必看篇),如下图

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

,见图

标致2008 漏油

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

标致2008 漏油ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

1.ajax调用返回php接口返回json数据的方法(必看篇)

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

2.ajax调用返回php接口返回json数据的方法(必看篇)。

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

3.ajax调用返回php接口返回json数据的方法(必看篇)。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

4.ajax调用返回php接口返回json数据的方法(必看篇)。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)ajax调用返回php接口返回json数据的方法(必看篇)

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

ajax调用返回php接口返回json数据的方法(必看篇)。标致2008 漏油

展开全文
相关文章
澳门红糖暖胃供应商

ajax调用返回php接口返回json数据的方法(必看篇)

大发国

ajax调用返回php接口返回json数据的方法(必看篇)....

人立方

ajax调用返回php接口返回json数据的方法(必看篇)....

二八论坛上不去了

ajax调用返回php接口返回json数据的方法(必看篇)....

广水三维动漫案例

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

....

相关资讯
怎样识别mg电子游戏

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

....

蒙特卡洛代理申请

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

....

澳门专业通风管道报价

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

....

方网彩票网上投注

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

....

长乐新生儿照背景模板

php代码如下:

<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; }?>

js代码如下:

<script> $.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) { // data = $.parseJSON(result); // alert(data.nickname); alert(result.nickname); } }); </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret['foo'] = "bar";finish();function finish() { header("content-type:application/json"); if ($_GET['callback']) { print $_GET['callback']."("; } print json_encode($GLOBALS['ret']); if ($_GET['callback']) { print ")"; } exit; }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

....

热门资讯