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网页设计。