ajax传递的问题

//////html页面

<table align="center" cellpadding="5" cellspacing="0" border="0" style="font-size:9pt;">
<tr>
     <td>用户名：</td>
        <td><input type="text" name="user_name" id="user_name" style="width:250px; height:18px;" /></td>
        <td><span id="td_user_name">请输入你想拥有的用户名</span></td>
    </tr>
<tr>
     <td>密码：</td>
        <td><input type="password" id="userpwd" name="userpwd" style="width:250px; height:18px;" /></td>
        <td><span id="td_userpwd">请牢记自己输入的密码</span></td>
    </tr>
    <tr>
     <td> </td>
        <td align="center" colspan="2"><input type="submit" id="submit" value="提交" />
       <input type="reset" id="reset" value="重置" /></td>
    </tr>
</table>

/////////////////////ajax

function creatXMLHTTP(){                                //创建XMLHttpRequest
  var request;
  var browser = navigator.appName;                //得到当前浏览器        
  if(browser == 'Microsoft Internet Explorer'){     //IE
    request = new ActiveXObject('Microsoft.XMLHTTP');
    return request;
   }else {                                      //非IE
     request = new XMLHttpRequset();
     return request;
    }
}

var xrh = creatXMLHTTP();

function for_check(){
   var url = "aa.php?user_name="+escape(document.getElementById('user_name').value);
   //var aaa = document.getElementById('user_name').value;
   //alert(aaa);
   xrh.open("GET",url,true);
   xrh.onreadystatechange = check;
   xrh.send(null);
  }
function check(){
   if(xrh.readystate == 4){
     var zzz = xrh.responseTest;
     /*
     if(zzz == "1"){
       document.getElementById('td_user_name').innerHTML = "已经被别人使用...";
      }else {
        document.getElementById('td_user_name').innerHTML = "可用用户名...";
       }
     */
       alert(zzz);
    }
  }



////php处理代码


$conn = mysql_connect($host,$username,$password);
mysql_select_db($database,$conn);
mysql_query("SET NAMES 'utf8'");
$user_name = iconv('UTF-8','gb2312',$_GET['user_name']);
//echo $user_name;
//$user_name = $_GET['user_name'];
$sql = "select user_name from user where user_name='".$user_name."'";
$rs = mysql_query($sql);
if (mysql_num_rows($rs) != 0){
///echo "1";
echo mysql_num_rows($rs);
}else {
echo "2";
}


问题出在哪里？为什么alert（zzz）得到是undefined？？？不管传递的是中文还是英文。。。。都是一样的请况，