ajax技术始终出去发送状态-php教程-PHP中文网

ajax技术始终出去发送状态

php中文网

发布： 2016-06-23 14:01:45

原创

860人浏览过

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
           if($id=$_GET[id]){
  for($i=1;$i
echo $i;
sleep(2);
   }
   exit();
}
?>
index.php代码：
          <script></script>
o
t
x

回复讨论(解决方案)

if($id=$_GET[id]){  for($i=1;$i<=5;$i++){echo $i;//sleep(2);   }   exit();}

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<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>

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我那个哪里出现问题了？

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

谢谢~~~~~~~~~~~