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利用AJAX实现鼠标悬浮获取值的代码

2022-2-5 17:00| 发布者: 菜鸟教程小白| 查看: 297| 评论: 0|原作者: [db:作者]|来自: [db:来源]

摘要: 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

 
<!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"; 
} 
?> 