1

在第一次访问时，Page_Load 正在运行并且会话存储正确。当我单击登录按钮时，再次调用 Page_Load，然后调用 btnSignIn_Click 函数，但 Session 为空！

public partial class LoginPage : System.Web.UI.Page
{
    UserItem userItem = null;
    protected void Page_Load(object sender, EventArgs e)
    {
        //Initialize and validate post
         RequestObj post = new RequestObj(Context);            
         if (post.isValid)
         {
            Session["post"] = post;
         }
    }
    protected void btnSignIn_Click(object sender, EventArgs e)
    {
        if (Session["post"] != null)
        {
            RequestObj post = Session["post"] as RequestObj;    
            userItem = Functions.LogIn(post);
        }
        Response.Redirect("LogIn.aspx");
     }
}

///////////////////////////////////////// ///////////////////////////////////////// ///////////////////////////////////////// ////////////

<%@ Page Language="C#" AutoEventWireup="true" CodeBehind=" LoginPage.aspx.cs"
         Inherits="myNameSpace.LoginPage " %>

<!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 runat="server">
<meta name="viewport" content="width=320"/>
    <title></title>
</head>
<body>
<form id="form1" runat="server">
    <div style="width:100%; text-align:center;">    
    <%if (userItem == null)
      {%>
           Username:
        <asp:TextBox ID="txtUsername" runat="server"></asp:TextBox>
        <br />
        Password:
        <asp:TextBox ID="txtPass" runat="server" TextMode="Password">
        </asp:TextBox>
        <br />
        <asp:Button ID="btnSignIn" runat="server" Text="Sign In" 
            onclick="btnSignIn_Click" />
      <%}else{%>
       <%=userItem.LoginName%>
       <br />
       <%=userItem.LoginTime.ToString()%>
       <br />
        <asp:Button ID="btnSignOut" runat="server" Text="Sign Out" 
            onclick="btnSignOut_Click" />
       <%}%>
    </div>
    </form>
</body>
</html>
4

1 回答 1

1

您的意思是仅在页面最初呈现时设置会话吗？

尝试像这样包装作业

if(!IsPostBack)
{
   RequestObj post = new RequestObj(Context);            
   if (post.isValid)
   {
     Session["post"] = post;
   }
}

这将防止它在回发期间被重置

否则我会在调试器中检查 post 的值

于 2013-07-03T05:33:40.017 回答