AJAX实现鼠标悬浮获取值

来源:互联网 发布:剑三妖娆御姐捏脸数据 编辑:程序博客网 时间:2024/06/03 23:19

 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

<?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";
}  

?>

原创粉丝点击