ajax调用返回php接口返回json数据的方法(必看篇)

Doris ·

更新时间:2024-09-21

· 956 次阅读

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。

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软件开发网。

您可能感兴趣的文章:原生js调用json方法总结json跨域调用python的方法详解JavaScript跨域调用基于JSON的RESTful API微信小程序使用wx.request请求服务器json数据并渲染到页面操作示例微信小程序通过api接口将json数据展现到小程序示例微信小程序学习(4)-系统配置app.json详解微信小程序如何调用json数据接口并解析

ajax调用 json数据 JSON 方法 PHP AJAX

1024 个赞