下面是我的完整代码,为什么点了o,x,t之后都没有反应呢,按道理应该是下面会多出一些内容的,大家知道为什么吗?var xmlHttp;
function xmlhttprequest() { if(window.ActiveXObject){
xmlHttp = new ActiveXObject('Microsoft.XMLHTTP');
}else if(window.XMLHttpRequest){
xmlHttp = new XMLHttpRequest();
}}function funphp100(url){
xmlhttprequest();
xmlHttp.open("GET","for.php?id="+url,"true");
xmlHttp.onreadystatechange = byphp;
xmlHttp.send(null);
}function byphp(){ if(xmlHttp.readyState==1){
document.getElementById('php100').innetHTML = "loading......";
} if(xmlHttp.readyState==4){
if(xmlHttp.Status==200){
var byphp100 = xmlHttp.responseText;
document.getElementById('php100').innetHTML = byphp100;
}
}
}
for.php
<?php
for($i=1;$i<=6;$i++){
echo $i."$id<br>";
}
?>
index.php<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>
function xmlhttprequest() { if(window.ActiveXObject){
xmlHttp = new ActiveXObject('Microsoft.XMLHTTP');
}else if(window.XMLHttpRequest){
xmlHttp = new XMLHttpRequest();
}}function funphp100(url){
xmlhttprequest();
xmlHttp.open("GET","for.php?id="+url,"true");
xmlHttp.onreadystatechange = byphp;
xmlHttp.send(null);
}function byphp(){ if(xmlHttp.readyState==1){
document.getElementById('php100').innetHTML = "loading......";
} if(xmlHttp.readyState==4){
if(xmlHttp.Status==200){
var byphp100 = xmlHttp.responseText;
document.getElementById('php100').innetHTML = byphp100;
}
}
}
for.php
<?php
for($i=1;$i<=6;$i++){
echo $i."$id<br>";
}
?>
index.php<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>
解决方案 »
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