<?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 addingcallback=?
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 likejQuery17209314005577471107_1335958194322({"foo":"bar"})
Here is the PHP code I used to achieve this,which degrades if JSON (without a callback) is used:
$ret['foo'] = "bar"; finish(); function(){ header("content-type:application/json");if($_GET'callback'])print $_GET]."("} json_encode$GLOBALS'ret']);")"exit}
Hopefully that will help someone in the future.
2.第二个问题:
VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1
原文地址:https://www.jb51.cc/ajax/162191.html
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