散分啦!!!!!!
怎么样在不刷新页面的情况下保存下密码??
[解决办法]
ajax
[解决办法]
ajax.asp.net下载ajax框架，利用里面的UpdatePanel控件实现
[解决办法]
//客户端登录页面 Login.aspx ;代码如下
<HTML>
<HEAD>
<title> HelloWorld </title>
<meta name= "GENERATOR " Content= "Microsoft Visual Studio .NET 7.1 ">
<meta name= "CODE_LANGUAGE " Content= "C# ">
<meta name= "vs_defaultClientScript " content= "JavaScript ">
<meta name= "vs_targetSchema " content= "http://schemas.microsoft.com/intellisense/ie5 ">
<script language= "jscript " src= "../Javascript/CallBackObject.js "> </script>
<script language= "jscript ">
function createRequest()
{
var name = escape(document.getElementById( "name ").value);
var pass = escape(document.getElementById( "pwd ").value);
var cbo = new CallBackObject();
cbo.OnComplete = Cbo_Complete;
cbo.onError = Cbo_Error;
cbo.DoCallBack( "Ajax.aspx?name= "+name+ "&password= "+pass);
}
function Cbo_Complete(responseText, responseXML)
{
alert(responseText);
}
function Cbo_Error(status, statusText, responseText)
{
alert(responseText);
}
</script>
</HEAD>
<body>
<form id= "Form1 " method= "post " runat= "server ">
<asp:Label id= "lblUsername " runat= "server " Width= "70 "> UserName: </asp:Label> <INPUT id= "name " style= "WIDTH: 192px; HEIGHT: 22px " type= "text " size= "26 ">
<br>
<asp:Label id= "lblPassword " runat= "server " Width= "70px "> Password: </asp:Label> <INPUT id= "pwd " style= "WIDTH: 192px; HEIGHT: 22px " type= "text " size= "26 ">
<br>
<input type= "button " value= "Login! " onclick= "createRequest() " style= "WIDTH: 128px; HEIGHT: 24px ">
</form>
</body>
</HTML>
服务器处理页面：AJAX.asp.cs
using System;
using System.Collections;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Web;
using System.Web.SessionState;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Web.UI.HtmlControls;
using System.Data.SqlClient;
namespace AjaxFrmClass
{
/// <summary>
/// AjaxTest 的摘要说明。
/// </summary>
public class AjaxTest : System.Web.UI.Page
{
private void Page_Load(object sender, System.EventArgs e)
{
//show();
#region 用户登陆
switch(LoginVerify())
{
case "1 ":
HttpContext.Current.Response.Clear();
HttpContext.Current.Response.Write( "密码不正确 ");
HttpContext.Current.Response.Flush();
HttpContext.Current.Response.End();
break;
case "2 ":
HttpContext.Current.Response.Clear();
HttpContext.Current.Response.Write( "用户名不存在 ");
HttpContext.Current.Response.Flush();
HttpContext.Current.Response.End();
break;
default:
break;
}
#endregion
}
#region 输入地用户名进行判断
/// <summary>
/// 对用户输入的用户名和密码进行有效性验证
/// </summary>
/// <returns> </returns>
public string LoginVerify()
{
try
{
string strName = HttpContext.Current.Request.QueryString[ "name "];
string strPwd = HttpContext.Current.Request.QueryString[ "password "];
SqlParameter MyParameter= new SqlParameter();
SqlCommand cmd = new SqlCommand();
SqlConnection con = new SqlConnection( "uid=sa;pwd=1234;database=pubs ");
cmd.CommandType = CommandType.StoredProcedure;
cmd.CommandText = "SP_User_Login ";
cmd.Connection = con;
MyParameter = cmd.Parameters.Add( "@RETURN_VALUE ",SqlDbType.Int);
MyParameter.Direction = ParameterDirection.ReturnValue;
cmd.Parameters.Add( "@name ",strName);
cmd.Parameters.Add( "@pwd ", strPwd);
con.Open();
cmd.ExecuteNonQuery();
return cmd.Parameters[ "@RETURN_VALUE "].Value.ToString();
}
catch(Exception errExce)
{
return errExce.Message.ToString();
}
finally
{
}
}
#endregion
#region Web 窗体设计器生成的代码
override protected void OnInit(EventArgs e)
{
//
// CODEGEN: 该调用是 ASP.NET Web 窗体设计器所必需的。
//
InitializeComponent();
base.OnInit(e);
}
/// <summary>
/// 设计器支持所需的方法 - 不要使用代码编辑器修改
/// 此方法的内容。
/// </summary>
private void InitializeComponent()
{
this.Load += new System.EventHandler(this.Page_Load);
}
#endregion
}
}
Ajax.aspx 不需要做处理，此时只需要在服务器端处理即可
完成以上2个页面的操作，基本上算是成功了，就差一个登录的存储过程了
CREATE PROC SP_User_Login
(
@name VARCHAR(15),
@pwd VARCHAR(15)
)
AS
DECLARE @temp_Name VARCHAR(15)
BEGIN
SELECT @temp_Name=UserName
FROM t_UserLogin
WHERE UserName = @name
AND
Password = @pwd
IF @temp_Name IS NOT NULL
RETURN 0 --登录成功
IF EXISTS (SELECT UserName
FROM t_UserLogin
WHERE UserName = @name)
RETURN 1 --密码不正确
ELSE
RETURN 2 --用户名不存在
END
GO
