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ajax实现将鼠标放到图标上，下方会显示和该图有关的信息
客户端代码mouseover.php
复制代码 代码如下:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
<title>鼠标悬浮测试</title>
</head>
<script type="text/javascript" language="javascript">
var xmlHttp;
function createXMLHttpRequest(){
if(window.ActiveXObject){
xmlHttp = new ActiveXObject("microsoft.XMLHTTP");
}
else if(window.XMLHttpRequest){
xmlHttp = new XMLHttpRequest();
}
}
function sendRequest(t){
var id = t.id
createXMLHttpRequest();
var url = "mouseover_check.php?page=" +t.id ;
xmlHttp.onreadystatechange = callback;
xmlHttp.open('GET',url,true);
xmlHttp.send(null);
}
function callback(){
if(xmlHttp.readyState == 4){
if(xmlHttp.status == 200){
document.getElementById("show").innerHTML = xmlHttp.responseText;
}
}
}
</script>
<body>
<p>
<input type="text" value="here" id="1" onmouseover="sendRequest(this)" />
</p>
<p><br/>
<input type="text" value="haha" id="2" onmouseover="sendRequest(this)" />
</div>
</p>
<p></p>
<span id="show"></span>
</body>
</html>
服务器端代码：
mouseover_check.php
复制代码 代码如下:
<?php
header("Content-type:text/html;charset=gb2312");
header("cache-control:no-cache,must-revalidate");
$name = $_GET['page'];
if($name == '1'){
echo "111";
}
else{
echo "222";
}
?>
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发表于 2009-12-4 13:04:37
