2014-07-02 60 views
2

我是新来的Ajax。我在一个名为home.php的文件中写入了两个函数,并对members.php进行了Ajax调用。在发送Ajax请求时得到响应的错误

我的问题是当我用ajax调用发送数据时,我无法获得响应。

home.php


<script language="javascript" type="text/javascript"> 

     var httpObject=false; 
     var httpObject1=false; 
     if(window.XMLHttpRequest){ 
      httpObject = new XMLHttpRequest(); 
      httpObject1 = new XMLHttpRequest(); 
     }else if(window.ActiveXObject){ 
      httpObject = new ActiveXObject("Microsoft.XMLHttp"); 
      httpObject1 = new ActiveXObject("Microsoft.XMLHttp"); 
     } 
     function logout(){ 

      if(httpObject.readyState == 4 && httpObject.status == 200){ 
        var response = httpObject.responseText; 

        if(response == "logout"){ 
         window.location.href="index.php"; 

        } 
        else{ 
         error.innerHTML = "Sorry, Invalid Login."; 
        } 
       } 
     } 

     function get_Request_Member_List(type){ 

      var member_Type = type; 
      var queryString = "?data1=" + member_Type ; 
      if(httpObject1.readyState == 4 && httpObject1.status == 200){ 
        var response = httpObject1.responseText; 
        document.getElementById(type).innerHTML = response; 
      } 

     } 

     httpObject.open("GET", "logout.php" ,true); 
     httpObject.send(null); 

     httpObject1.open("GET", "members.php"+queryString ,true); 
     httpObject1.send(null); 
    </script> 

members.php


<?php 
echo "hello"; 
?> 
+0

如何调用logout()和get_Request_Member_List()函数? – Phantom

+0

我这样做 –

+0

将您的open()和send()行移动到此函数。在此行后面:var queryString = ... – Phantom

回答

0

不要忘了一个onreadystatechange事件处理程序添加到您的XMLHTTPRequest的,如:

httpObject.onreadystatechange = logout; 
+1

如果请求是同步的,则这不是必需的。 – Phantom

+0

我把这个和为什么? –

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