Покажите сообщение об ошибке и предотвратите дублирование голосования

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protected void LinkButton_Click(Object sender, EventArgs e)
        {

            String MyConnection2 = "Server=localhost;database=ovs;Uid=root;password=; Convert Zero Datetime=True";
            DateTime time = DateTime.Now;              // Use current time
            string format = "yyyy-MM-dd HH:mm:ss";
            string UserName4 = HttpContext.Current.User.Identity.Name;
            GridViewRow grdrow = (GridViewRow)((LinkButton)sender).NamingContainer;
            Label lblStudentId = (Label)grdrow.Cells[0].FindControl("lblID");
          Label lblvID = (Label)GridView2.Rows[0].FindControl("lblvID");
           string label1val = lblvID.Text;

            string studentId = lblStudentId.Text;
            //    string id =;
            //      string testing = this.Encrypt(id.Trim());
            String query = "insert into voting (CandidateStudentID,voterStudentID,DateTime)values ('" + lblStudentId.Text + "','" + Session["UserName"].ToString() + "','" + time.ToString(format) + "')";
           
            if (Session["UserName"].ToString().Equals(lblvID.Text))
            {

                Label1.Text = "You voted beofre";

            }
            else
            {
                MySqlConnection MyConn2 = new MySqlConnection(MyConnection2);
                MySqlCommand MyCommand2 = new MySqlCommand(query, MyConn2);
                MySqlDataReader MyReader2;
                MyConn2.Open();
                MyReader2 = MyCommand2.ExecuteReader();

                Label2.Text = "Thank you for You Vote";


            }


        }


<pre><asp:GridView ID="GridView2" runat="server"  AutoGenerateColumns="False" Font-Size="Medium">
              <Columns>
              <asp:TemplateField HeaderText="Student ID">
      <ItemTemplate>
         <asp:Label ID="lblvID" runat="server"   Width="150px"  Text='<%#Eval("voterStudentID") %>'/>
     </ItemTemplate>
 </asp:TemplateField>

              </Columns>
           </asp:GridView>

Что я уже пробовал:

Кнопка link принадлежит gridview 1.я получаю ошибку index out of range в окне

Label lblvID = (Label)GridView2.Rows[0].FindControl("lblvID");.

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