Access a control inside a the LayoutTemplate of a ListView

a 夏天 提交于 2019-11-29 05:28:56

Try this:

((Literal)lv.FindControl("litControlTitle")).Text = "Your text";

The complete solution:

<asp:ListView ID="lv" OnLayoutCreated="OnLayoutCreated" runat="server">
  <LayoutTemplate>
    <asp:Literal ID="lt_Title" runat="server" />
    <asp:PlaceHolder ID="itemPlaceHolder" runat="server" />
  </LayoutTemplate>
  <ItemTemplate>
  </ItemTemplate>
</asp:ListView>

In codebehind:

protected void OnLayoutCreated(object sender, EventArgs e)
{
    (lv.FindControl("lt_Title") as Literal).Text = "Your text";
}

This technique works for template layout; use the init event of the control:

<asp:ListView ID="lv" runat="server" OnDataBound="lv_DataBound">
  <LayoutTemplate>
    <asp:Literal ID="litControlTitle" runat="server" OnInit="litControlTitle_Init" />
    <asp:PlaceHolder ID="itemPlaceHolder" runat="server" />
  </LayoutTemplate>
  <ItemTemplate>
  </ItemTemplate>
</asp:ListView>

And capture a reference to the control for use in the code-behind (e.g) in the ListView's DataBound event:

private Literal litControlTitle;

protected void litControlTitle_Init(object sender, EventArgs e)
{
    litControlTitle = (Literal) sender;
}

protected void lv_DataBound(object sender, EventArgs e)
{
    litControlTitle.Text = "Title...";
}
Dheeraj Palagiri

For Nested LV Loop:

void lvSecondLevel_LayoutCreated(object sender, EventArgs e)
{
    Literal litText = lvFirstLevel.FindControl("lvSecondLevel").FindControl("litText") as Literal;
    litMainMenuText.Text = "This is test";
}

In case you need the VB version, here it is

Dim litControl = CType(lv.FindControl("litControlTitle"), Literal)
litControl.Text = "your text"
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