ajax调用返回php接口返回json数据的方法(必看篇)

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=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。

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