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ajax技术 始终出去发送状态
- WBOYOriginal
- 2016-06-23 14:01:45783browse
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();} <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> 我那个哪里出现问题了？
else if( widow.XMLHttpRequest){
xmlHttp=new XMLHttpRequest();
}
window
谢谢~~~~~~~~~~~