How to assign UpdatePanel Trigger's control ID with button's from gridview

Question

currently I have a UpdatePanel for jQuery Dialog use, which contains a GridView.

And that GridView contains a FileUpload control in footer and EmptyDataTemplate

Answer 1

Try registering the post back control from code behind like this:

protected void grdExpense_RowCreated(object sender, GridViewRowEventArgs e)
    {
        LinkButton btnAdd = (LinkButton)e.Row.Cells[0].FindControl("btnAdd");
        if (btnAdd != null)
        {
            ScriptManager1.RegisterAsyncPostBackControl(btnAdd);
        }
    }

Answer 2

This works

protected void grdExpense_RowCreated(object sender, GridViewRowEventArgs e)
    {
        LinkButton btnAdd = (LinkButton)e.Row.Cells[0].FindControl("btnAdd");
        if (btnAdd != null)
        {
            ScriptManager.GetCurrent(this).RegisterPostBackControl(btnAdd);
        }

    }

Answer 3

I had a similar problem and this post helped me, but I found that registering the control in the scriptmanager works only if the updatepanels UpdateMode is set to "Always". If its set to "Conditional" this approach does not work.

I found another approach that always works which is to add triggers to the updatepanel in the DataBound() event of the gridview:

    Dim CheckBoxTrigger As UpdatePanelControlTrigger = New AsyncPostBackTrigger()
    Dim SelectCheckBox As CheckBox
    For i = 0 To GridViewEquipment.Rows.Count - 1 Step 1
        SelectCheckBox = GridViewEquipment.Rows(i).Cells(12).FindControl("CheckBoxSign")
        CheckBoxTrigger.ControlID = SelectCheckBox.UniqueID
        UpdatePanelEquipment.Triggers.Add(CheckBoxTrigger)
    Next

Answer 4

Instead of adding a trigger to upnlEditExpense maybe you can try to add an update panel around the link button inside the template with no triggers...

<asp:TemplateField>
     <FooterTemplate>
          <asp:UpdatePanel ID="upnlBtnAdd" runat="server">
              <ContentTemplate>
                    <asp:LinkButton runat="server" ID="btnAdd" Text="Add" OnClick="btnAdd_Click"></asp:LinkButton>
              </ContentTemplate>
          </asp:UpdatePanel>
     </FooterTemplate>
</asp:TemplateField>