请问大师们,查询注册用户名已存在时按钮不提交，不跳转，如何实现

// JavaScript Document
var XHR;
function createXHR(){
    if(window.ActiveXObject){
        XHR=new ActiveXObject('Microsoft.XMLHTTP');
    }else if(window.XMLHttpRequest){
        XHR=new XMLHttpRequest();
    }
}
function checkname(){
    var username=document.form1.host.value;
    createXHR();    
    XHR.open("GET","./checkname.php?id="+username,true);
    XHR.onreadystatechange=byhongfei;
    XHR.send(null);
}
function byhongfei(){
    if(XHR.readyState == 4){
        if(XHR.status == 200){    
            var textHTML=XHR.responseText;            
            document.getElementById('msg').innerHTML=textHTML;
        }
    }
}

用户名可以使用";
    }
mysql_close();
?>

document.getElementById("button").disabled    = 'true';//禁用
document.getElementById("button").removeAttribute("disabled");//恢复

------解决方案--------------------

JScript code

function byhongfei(){
    if(XHR.readyState == 4){
        if(XHR.status == 200){    
            var textHTML=XHR.responseText;            
            document.getElementById('msg').innerHTML=textHTML;
            if(textHTML == "用户名已存在"){
              document.getElementById("button").disabled    = 'true';//禁用
            }
        }
    }
}