AJAX 中怎么实现二级联级菜单，相信很多没有经验的人对此束手无策，为此本文总结了问题出现的原因和解决方法，通过这篇文章希望你能解决这个问题。
客户端代码：
<!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>无文档</title>
</head>
<script type="text/javascript">
var xmlHttp;
var a =new Array();
function createXMLHttpRequest(){
if(window.ActiveXObject){
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
else if(window.XMLHttpRequest){
xmlHttp=new XMLHttpRequest();
}
}
function send_request(){
createXMLHttpRequest();
var year = document.getElementById("year");
var url = "check_2.php?page="+escape(year.value);
xmlHttp.open("GET",url,true);
xmlHttp.onreadystatechange = checkit;
xmlHttp.send(null);
}
function checkit(){
if(xmlHttp.readystate == 4){
if(xmlHttp.status == 200){
showChild();
}
}
}
function showChild(){
var xmlDoc=xmlHttp.responseXML;
var content=xmlDoc.getElementsByTagName("city");
for(var i=0;i<content.length;i++){
var y=content[i];
a[i]=y.childNodes[0].data;
}
show1();
}
function show1(){
var obj=document.getElementById("name");
var number=obj.length;
for(var j=obj.length-1;j>=0;j--){
obj.removeChild(obj.childNodes.item(j));
}
for(var i=0;i<a.length;i++){
var opt=document.createElement("OPTION");
opt.text=a[i];
obj.add(opt);
}
}
</script>
<body>
年份：<select id="year" onchange="send_request()">
<option value="0">请选择</option>
<option value="1">1996-2006</option>
<option value="2">1986-1995</option>
<option value="3">1971-1985</option>
<option value="4">1970以前</option>
</select>
子目录:<select id="name">
<option value="0">请选择</option>
</select>
</body>
</html>
服务器端代码：
<?php
header('Content-type: text/xml');
$xml="<?xml version='1.0' encoding='GB2312'?>";
$year = $_GET["page"];
$content = $xml."<contents>";
if($year == "1"){
$content = $content."<city>1</city><city>11</city></contents>";
}
else if($year == '2'){
$content = $content."<city>2</city><city>12</city></contents>";
}
else if($year == '3'){
$content = $content."<city>3</city><city>13</city></contents>";
}
else if($year == '4'){
$content = $content."<city>4</city><city>14</city></contents>";
}
echo $content;
?>
看完上述内容，你们掌握AJAX 中怎么实现二级联级菜单的方法了吗？如果还想学到更多技能或想了解更多相关内容，欢迎关注编程网行业资讯频道，感谢各位的阅读！
