ajax技术始终出去发送状态_H5之家 - 中国HTML5教程资源分享第一站

ajax.js代码：
       var xmlHttp;
function S_xmlhttprequest(){
    if(window.ActiveXObject){
       xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
    else if(widow.XMLHttpRequest){
       xmlHttp=new XMLHttpRequest();
    }
}
function funphp100(url){
    S_xmlhttprequest();
    xmlHttp.open("GET","for.php?id="+url,true);
    xmlHttp.onreadystatechange=byphp;
    xmlHttp.send(null);
}
function byphp(){
   if(xmlHttp.readyState==1){
   document.getElementById('php100').innerHTML="loading....";
   }
   if(xmlHttp.readyState==4){

          var byphp100 = xmlHttp.responseText;
          document.getElementById('php100').innerHTML=byphp100;

    }

}
for.php代码
          <?php
if($id=$_GET[id]){
  for($i=1;$i<=5;$i++){

echo $i;
sleep(2);
   }
   exit();
}
?>
index.php代码：
          <script type="text/javascript" src= "ajax.js"></script>
<a href= "#" onclick="funphp100('o')">o</a>
<a href= "#" onclick="funphp100('t')">t</a>
<a href= "#" onclick="funphp100('x')">x</a>
<br>
<div id="php100"></div>

回复讨论(解决方案)

if($id=$_GET[id]){ for($i=1;$i<=5;$i++){ echo $i; //sleep(2); } exit(); }
<html> <head> <title> </title> <script type="text/javascript" src='jquery-1.4.2.min.js'></script> </head> <script type="text/javascript"> var xmlHttp; function S_xmlhttprequest(){ if(window.ActiveXObject){ xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); } else if(window.XMLHttpRequest){ xmlHttp=new XMLHttpRequest(); } } function funphp100(url){ S_xmlhttprequest(); xmlHttp.open("GET","test.php?id="+url,true); xmlHttp.onreadystatechange=byphp; xmlHttp.send(null); } function byphp(){ if(xmlHttp.readyState==1){ document.getElementById('php100').innerHTML="loading...."; } if(xmlHttp.readyState==4){ var byphp100 = xmlHttp.responseText; document.getElementById('php100').innerHTML=byphp100; } } </script> <body> <a href= "#" onclick="funphp100('o')">o</a> <a href= "#" onclick="funphp100('t')">t</a> <a href= "#" onclick="funphp100('x')">x</a> <br> <div id="php100"></div> </body> </html>

演示效果图片0

我那个哪里出现问题了？

else if( widow.XMLHttpRequest){
xmlHttp=new XMLHttpRequest();
}
window

谢谢~~~~~~~~~~~

ajax技术 始终出去发送状态

ajax技术始终出去发送状态