The state information is invalid for this page and might be corrupted. error

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The state information is invalid for this page and might be corrupted. Error occurs when click gridview events.this error occurs only when page load call display function.

C#

protected void Page_Load(object sender, EventArgs e)
    {
        string userID = Session["Member"].ToString();

       

       Display();
    }

protected void Display()
    {
        DataSet ds = new DataSet();
        string _UserName = Session["Member"].ToString();
        MemberUpload objDisplay = new MemberUpload();
        ds = objDisplay.GetUploadDetailsByUserName(_UserName, "D", "%");
        if (ds.Tables.Count > 0)
        {
            GvNews.DataSource = ds;
            GvNews.DataBind();
        }
        else
        {
            GvNews.DataSource = null;
            GvNews.DataBind();
        }
}

Posted 10-Sep-12 21:26pm

Shibiny

Updated 10-Sep-12 21:33pm

Sandeep Mewara

v2

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Ankur^\m/ · Answer 1 · 2012-09-10T21:34:00

Solution 1

Do some initial research. This should help - The state information is invalid for this page and might be corrupted[^]

If it doesn't help, come back again with your findings.

Posted 10-Sep-12 21:34pm

Ankur^\m/

Sandeep Mewara · Answer 2 · 2012-09-10T21:38:00

Solution 2

Use Page.IsPostback property. On event clicks, when page gets postback, your Display() method re-executes and attaches data with Grid.

Try:

C#

protected void Page_Load(object sender, EventArgs e)
{
   if(!IsPostback)
   {
     string userID = Session["Member"].ToString();
     Display();
   }
}

Posted 10-Sep-12 21:38pm

Sandeep Mewara

Comments

Shibiny 11-Sep-12 4:37am

yes.there is no error now.but i want it in ispostback