ajax调用返回php接口返回json数据的方法(必看篇)

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php代码如下：

<?php  

header('Content-Type: application/json');  

header('Content-Type: text/html;');  

$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/[email protected]",      

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"})</p><p>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。