怎么用AJAX來判斷dedecms用戶是否登錄呢

JS代碼：
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<script language="javascript" src="{dede:global name='cfg_cmspath'/}/include/dedeajax2.js"></script>
<script language="javascript">
function CheckUser(){
var taget_obj = document.getElementById('_userboxform');
myajax = new DedeAjax(taget_obj,false,false,"","","");
myajax.SendGet2("{dede:global name="cfg_memberurl"/}/userboxsta.php");
DedeXHTTP = null;
}
</script>
然后我們修改下內容頁部分的代碼
找到：
<div class="userbox">
<dl>
<dd class="mtop">
<strong>用戶名：</strong><input name="username" class="username" type="text" maxlength="20" />
<strong>密碼：</strong><input name="pwd" class="password" type="password" maxlength="20" />
</dd>
將其修改為
<div class="userbox">
<dl>
<dd class="mtop"> <span id="_userboxform">
<strong>用戶名：</strong><input name="username" class="username" type="text" maxlength="20" />
<strong>密碼：</strong><input name="pwd" class="password" type="password" maxlength="20" /> </span> 
</dd>
即加上<span id="_userboxform">，用於判斷用戶是否已經登錄。
下面我們來寫一�php文件，用戶傳送數據，將其保存為/member/userboxsta.php,其代碼如下：
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<?php 
header("Pragma:no-cache\r\n");
header("Cache-Control:no-cache\r\n");
header("Expires:0\r\n");
header("Content-Type: text/html; charset=gb2312");
//系統設置為維護狀態可訪問
$cfg_IsCanView = true;
require_once(dirname(__FILE__)."/../include/inc_memberlogin.php");
$cfg_ml = new MemberLogin(); 
if(empty($cfg_ml->M_ID)){ echo ""; exit(); }
$uid = $cfg_ml->M_LoginID;
?>
<strong>歡迎您</strong>[<?php echo $cfg_ml->M_UserName?>],現在你可以直接發布留言