股利没有得到由AJAX请求填充

Question

我试图做一个与jQuery的ajax调用，它似乎是工作，但我不能让它填充结果股利。

继承人的JS代码：

  <script> 
       $(document).ready(function(){ 
        $(document).on('change', '#flip-1', function(){  
         var datastring = $("#some-form").serialize(); 
         $.ajax 
         ({ 
          type: "POST", 
          url: "test.php", 
          data: datastring, 
          dataType: 'html', 
          complete: function(data) { 
           $('#results').html (data); 
           } 
          }); 
         }); 
        }); 
      </script> 
      <form id="some-form"> 
       <label for="flip-1">Flip toggle switch:</label> 
       <select id="flip-1" name="flip-1" data-role="flipswitch"> 
        <option value="off">Off</option> 
        <option value="on">On</option> 
       </select> 
      </form> 

<div id="results">Loading users...</div> 

这里是PHP代码：

<?php 

if ($_POST['flip-1'] == 'on') { 
    echo 'oh is on!'; 
} elseif ($_POST['flip-1'] == 'off') { 
    echo 'no, its not on'; 
} 
?>  

这就是铬 “检查元素” 调试说一下头：

Request URL:http://localhost/smart_home/test.php 
Request Method:POST 
Status Code:200 OK 

Request Headers 
POST /smart_home/test.php HTTP/1.1 
Host: localhost 
Connection: keep-alive 
Content-Length: 9 
Accept: text/html, */*; q=0.01 
Origin: http://localhost 
X-Requested-With: XMLHttpRequest 
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/33.0.1750.117 Safari/537.36 
Content-Type: application/x-www-form-urlencoded; charset=UTF-8 
Referer: http://localhost/smart_home/index.php 
Accept-Encoding: gzip,deflate,sdch 
Accept-Language: sv-SE,sv;q=0.8,en-US;q=0.6,en;q=0.4 

Form Data 
flip-1=on 

Response Headers 
HTTP/1.1 200 OK 
Date: Thu, 27 Feb 2014 20:58:08 GMT 
Server: Apache/2.4.7 (Win32) OpenSSL/1.0.1e PHP/5.5.6 
X-Powered-By: PHP/5.5.6 
Content-Length: 20 
Keep-Alive: timeout=5, max=99 
Connection: Keep-Alive 
Content-Type: text/html 

这就是响应选项卡所说的： 当按钮变成on posit离子：

oh is on!  

时更改为 “关”

no, its not on  

没有任何人有什么想法？

谢谢！

Answer 1

complete回调没有直接接收响应数据（第一个参数是​​对象）。这是一般简单的使用success或.done()回调，而不是：

$.ajax({ 
    type: "POST", 
    url: "test.php", 
    data: datastring, 
    dataType: 'html', 
    success: function (data) { 
     $('#results').html(data); 
    } 
}); 

或者

$.ajax({ 
    type: "POST", 
    url: "test.php", 
    data: datastring, 
    dataType: 'html' 
}).done(function (data) { 
    $('#results').html(data); 
}); 

见http://api.jquery.com/jQuery.ajax/