<?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。
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