ajax传递一个参数具体实现
            网络编程 发布日期:2025/11/4 浏览次数:1
         
        
            正在浏览:ajax传递一个参数具体实现
            复制代码 代码如下: 
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<title></title> 
<script src="/UploadFiles/2021-04-02/Jquery1.7.js"><script type="text/javascript"> 
$(function () { 
$('#txtUserName').blur(function () { 
$.ajax({ 
type: "post", 
contentType: "application/json", 
url: "WebService1.asmx/ValidateUser", 
data: "{username:'" + $('#txtUserName').val() + "'}", 
success: function (result) { 
$('#spinfo').text(result.d); 
} 
}) 
}) 
}) 
</script> 
</head> 
<body> 
<div> 
用户名<input type="text" id="txtUserName" /><span id="spinfo"></span><br /> 
密码<input type="text" id="txtPwd" /> 
</div> 
</body> 
</html>