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...
<input name="LoginName" type="text" onblur="javascript:useradd()"/>
...
<script language="javascript">
function useradd()
{
var http_request = null;
function send_request(url){//初始化,发送请求函数
if (window.ActiveXObject) {
http_request = new ActiveXObject("Microsoft.XMLHTTP");
}
else if (window.XMLHttpRequest) {
http_request=new XMLHttpRequest();
}
http_request.onreadystatechange=processRequest;
http_request.open("GET",url,true);
http_request.send(null);
}
function processRequest()
{
if(http_request.readyState == 4)//判断对象状态
{
if(http_request.status == 200)
{
var error = http_request.responseText;
var rtstr = trim(error);
if (rtstr.length > 0)
{
alert('该用户名已存在!');
document.getElementById('addbtn').disabled = "true";
return;
}
else
{
document.getElementById('addbtn').disabled = null;
}
}
else
{
alert("Unsuccessful");
}
}
}
if(userAddForm.LoginName.value.length == 0)
return;
var vName="";
if (window.ActiveXObject) {
vName = escape(userAddForm.LoginName.value,'utf-8');
}
else
{
vName = escape(userAddForm.LoginName.value);
}
send_request("/xf_sms/Home/User/checkUserName.jsp?userName="+vName);
}
</script>
1、有没有进入后台的方法,进入了后台有没有13这个数据
2、send_request("/xf_sms/Home/User/checkUserName.jsp?userName="+vName);
在之前alert(vName)看看是否是13,而不是非法字符
3、你好像把ajax方法写在一个方法中了,这样不利于重用ajax方法