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

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发布： 2016-06-13 13:34:51

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请教大师们,查询注册用户名已存在时按钮不提交，不跳转，怎么实现？

PHP code

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JScript code

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// 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;
        }
    }
}

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PHP code

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<?php mysql_connect("localhost",'root','123456');
    mysql_select_db('dodiscuz_freebb');
    $sql="select * from member where username='$_GET[id]'";
    $query=mysql_query($sql);
    if(is_array(mysql_fetch_array($query))){
        echo "用户名已存在";
    }else{
        echo "<font color=green>用户名可以使用";
    }
mysql_close();
?>

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用户名存在时注册按钮怎样实现不可用状态现在就是用户名存在也跳转到index.php?reg

------解决方案--------------------

JScript code


document.getElementById("button").disabled    = 'true';//禁用
document.getElementById("button").removeAttribute("disabled");//恢复
<br><font color="#e78608">------解决方案--------------------</font><br>

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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
            }
        }
    }
} <div class="clear">
                 
              
              
        
            </div>

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