Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • Home
  • SEARCH
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 5840487
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 22, 20262026-05-22T11:45:12+00:00 2026-05-22T11:45:12+00:00

I have in many places in my ASP.NET project used the Session variable for

  • 0

I have in many places in my ASP.NET project used the Session variable for storing data. I usually write something like this:

public uint MyPropery 
{
    get 
    {
        object o = Session["MyProperty"];
        if (o != null)
            return (uint)o;
        else
            return 0;
    }
    set 
    {
        Session["MyProperty"] = value;
    }
}

However, this time I get a NullReferenceException in the setter. As far as I know, it is valid to assign the Session variable in the manner above. Also, Session is not null and neither is value.

Any ideas on this?

Edit:

Adding the code for the UserControl in which the property exists. I am using ext.net but that shouldn’t have anything to do with this. One thought that crossed my mind:

The UserControl (seen below) is added dynamically in code-behind of a page. Can that have anything to do with it?

I am adding UserControls like this (on a Page):

foreach(CoreCommons.System.Comment c in cg.Reply_Comments)
{
    WebApplicationExtNetTest.Secure.UserControls.CoreComment cc = new UserControls.CoreComment();
    cc._Comment = c; // here is where i get the NullRef
    this.Panel1.ContentControls.Add(cc);
}

Markup:

<%@ Control Language="C#" AutoEventWireup="true" CodeBehind="CoreComment.ascx.cs" Inherits="WebApplicationExtNetTest.Secure.UserControls.CoreComment" %>
<%@ Register Assembly="Ext.Net" Namespace="Ext.Net" TagPrefix="ext" %>

<ext:Panel runat="server" ID="CoreCommentOuterPanel" BodyStyle="background: #FFFDDE">
    <Items>
        <ext:ColumnLayout runat="server">
            <Columns>
                <ext:LayoutColumn ColumnWidth="0.8">
                    <ext:Image runat="server" ImageUrl="/Resources/bullet_triangle_green_16x16.png" Align="AbsMiddle"></ext:Image> 
                    <ext:Label runat="server" ID="lblCommentInfo"></ext:Label>
                </ext:LayoutColumn>
                <ext:LayoutColumn ColumnWidth="0.2"><ext:Button runat="server" ID="btnDelete" Icon="Delete"></ext:Button></ext:LayoutColumn>
            </Columns>
        </ext:ColumnLayout>
        <ext:Label runat="server" ID="lblComment"></ext:Label>
    </Items>
</ext:Panel>

Code-behind:

namespace WebApplicationExtNetTest.Secure.UserControls
{
    public partial class CoreComment : System.Web.UI.UserControl
    {
        public CoreCommons.System.Comment _Comment
        {
            get
            {
                object o = Session["CoreComment_ObjectId"];
                if (o != null)
                    return (tWorks.Core.CoreCommons.System.Comment)o;
                else
                    return null;
            }
            set
            {
                Session["CoreComment_ObjectId"] = value;
                SetComment();
            }
        }            

        protected void Page_Load(object sender, EventArgs e)
        {    
        }

        private void SetComment()
        {
            if (_Comment == null)
            {
                lblCommentInfo.Text = "";
                lblComment.Text = "";
            }
            else
            {
                lblCommentInfo.Text = _Comment.Author + ", " + _Comment.TimeStamp.ToString("g");
                lblComment.Text = _Comment.Text;
            }
        }
    }
}
  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-22T11:45:13+00:00Added an answer on May 22, 2026 at 11:45 am

    I’m almost completely sure the NullReferenceException is thrown in SetComment() because none of the CoreComment‘s child controls (lblComment, lblCommentInfo) are properly instantiated at the point you set the _Comment property.

    The reason these child controls are not instantiated is indeed the way you currently add the CoreComment controls. For dynamically adding UserControls, you must use Page.LoadControl() (see: here) to create a new instance of the control, as it does some behind-the-scenes magic to ensure it is properly initialized, which includes the instantiation of the child controls.

    On a sidenote, personally I’d change SetComment() to SetComment(CoreCommons.System.Comment comment) and use the parameter instead of repeatedly calling the getter, or, if staying with the original, at least call the getter only once and store the result in a local variable. With what I assume is probably InProc session storage it won’t make much of a difference, but in any other storage mode you’d repeatedly deserialize the Comment object for no reason.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

No related questions found

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.