<"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).")";
  }

"background-color: #ccffcc">js代码如下：



<script>
    $.ajax({
      url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php",
      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="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。