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ajax技术 始终出去发送状态,该如何解决
- WBOYOriginal
- 2016-06-13 11:49:24734browse
ajax技术 始终出去发送状态
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 $i;
sleep(2);
}
exit();
}
?>
index.php代码：
<script></script>
o
t
x
------解决方案--------------------
if($id=$_GET[id]){<br /> for($i=1;$i<=5;$i++){<br /><br />echo $i;<br />//sleep(2);<br /> }<br /> exit();<br />}<html><br /><head><br /> <title> </title> <br /> <script type="text/javascript" src='jquery-1.4.2.min.js'></script><br /><br /></head><br /><script type="text/javascript"><br /> var xmlHttp;<br /> function S_xmlhttprequest(){<br /> if(window.ActiveXObject){<br /> xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");<br /> }<br /> else if(window.XMLHttpRequest){<br /> xmlHttp=new XMLHttpRequest();<br /> }<br /> }<br /> function funphp100(url){<br /> S_xmlhttprequest();<br /> xmlHttp.open("GET","test.php?id="+url,true);<br /> xmlHttp.onreadystatechange=byphp;<br /> xmlHttp.send(null);<br /> }<br /> function byphp(){<br /> if(xmlHttp.readyState==1){<br /> document.getElementById('php100').innerHTML="loading....";<br /> }<br /> if(xmlHttp.readyState==4){<br /><br /> var byphp100 = xmlHttp.responseText;<br /> document.getElementById('php100').innerHTML=byphp100;<br /><br /> }<br /><br /> }<br /><br /></script><br /><body><br /> <a href= "#" onclick="funphp100('o')">o</a><br /> <a href= "#" onclick="funphp100('t')">t</a><br /> <a href= "#" onclick="funphp100('x')">x</a><br /> <br><br /> <div id="php100"></div><br /></body><br /></html>
------解决方案--------------------
else if(widow.XMLHttpRequest){
xmlHttp=new XMLHttpRequest();
}
window