ajax调用返回php接口返回json数据

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=[e-mail]",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:

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。