jquery发送ajax请求返回处理函数提醒缺少对象错误，求解！！！ - 编程论坛

HTML代码：
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.
<html xmlns="http://www. xml:lang="zh" lang="zh" dir="ltr">
<head>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Test</title>
<script type="text/javascript" src="./jquery.js"></script>
<script type="text/javascript" src="./test.js"></script>
</head>
<body>
<ul>
    <li><a href="javascript:viod(0);" onclick="Navigation('home')">首页</a></li>
    <li><a href="javascript:viod(0);" onclick="Navigation('anli')">案例</a></li>
</ul>
<div id="home"></div>
<div id="anli"></div>
</body>
</html>

js代码：
function Navigation(eid) {
    obj = $("#"+eid);
    $.ajax({
        type: "POST",
        url: "test.php",
        data:'key='+eid,
        Type: 'text',
        timeout: 5000,
        beforeSend: function(){
            obj.html('请求中....');
        },
        error : function(msg){
            obj.html('请求失败！');
        },
        success: function(msg){
            obj.html(msg);
        }
    });
}

php处理代码：
<?php
if('home'==$_POST['key']){
    print_r('<font color=red><h1>HOME</h1></font>');
}
if('anli'==$_POST['key']){
    print_r('<font color=red><h1>ANLI</h1></font>');
}
?>

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