ajax调用返回php接口返回json数据ajaxjsonpajaxjson实例ajax获取后台json数

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).")";
    }
?>

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.

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。

以上就介绍了ajax调用返回php接口返回json数据，包括了ajax,json方面的内容，希望对PHP教程有兴趣的朋友有所帮助。