ajax實現調用返回php介面返回json資料的方法

本篇文章主要介紹ajax實現調用返回php介面返回json資料的方法，感興趣的朋友參考下，希望對大家有所協助。

php代碼如下：

<?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代碼如下：

<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>

其中遇到了兩個問題：

1、第一個問題：

Uncaught SyntaxError: Unexpected token :

解決方案如下：

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、第二個問題：

解析json資料。從上面的javascript中可以看到，我沒有使用jquery.parseJSON()這些方法，開始使用這些方法，但是總是會報

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的錯誤，後來不用jquery.parseJSON()這個方法，反而一切正常。不知為何。