注册页面Ajax检查用户名重复,该如何解决

注册页面Ajax检查用户名重复
注册页面Ajax检查用户名重复

IE现在60行报错：'null'为空或不是对象

上代码：

HTML code

<!--

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-->    <script type="text/javascript">
        var xmlHttp;
        function createXMLHTTP()
        {
            if(window.ActiveXObject)
            {
                var browseVersion=["Msxml2.XMLHTTP.6.0","Msxml2.XMLHTTP.3.0","Msxml2.XMLHTTP","Microsoft.XMLHTTP"];
                for(var i=0;i<browseVersion.length;i++)
                {
                    xmlHttp=new ActiveXObject(browseVersion[i]);
                }   
            }
            else if(window.XMLHttpRequest)
            {
                xmlHttp=new XMLHttpRequest();
            }
            if (!xmlHttp)
            {
                window.alert("不能创建XMLHttpRequest对象实例！");
                return false;
            }
        }
        function $(id)
        {
            return document.getElementById(id);
        }
        function CheckUserName(obj,Msg)
        {
       
            if(obj.value=="")
            {
                alert( "请输入用户名!")
            }
            else
            { 
                createXMLHTTP();
                var url="chkuser.php?name=" + encodeURIComponent(obj.value);
                xmlHttp.open("GET",url,true);
                xmlHttp.onreadystatechange=IsExistUserName;
                xmlHttp.send(null);
            }
        }
        function IsExistUserName()
        {
            if(xmlHttp.readyState == 4)　　/*这是60行*/
            {
                var isValid = xmlHttp.responseText;
                var exsits = document.getElementById("exsits"); 
                exsits.innerHTML = isValid.substring(0,4);
            } 
         }
    </script>    

增加新用户
用户名：	*
密码：	*
重复密码：	*

这是chkuser.php页面代码：

PHP code

<!--

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--><?php include('chek.php');
include('../config.php');
$name = $_GET[name];
$sql="SELECT * FROM wyx_user WHERE wyx_name='$name'";
$sql=@mysql_query($sql)
if($rs=mysql_fetch_array($sql))
{
    echo "用户名已存在";
}
else
{
    echo "可以注册";
}

?>

------解决方案--------------------