Setting a WPF ContextMenu's PlacementTarget property in XAML?

Answer 1

You should be setting the ContextMenuService.Placement attached property on the button, as stated in the remarks in the documentation for ContextMenu.Placement.

<Button Name="btnFoo" Content="Foo" ContextMenuService.Placement="Bottom">
    <Button.ContextMenu>
        <ContextMenu>
            <MenuItem Header="Bar" />
        </ContextMenu>
    </Button.ContextMenu>
</Button>

Answer 2

Have you tried this:

<Button Name="btnFoo" Content="Foo">
    <Button.ContextMenu>
        <ContextMenu>
            <MenuItem Header="Bar" />
        </ContextMenu>
    </Button.ContextMenu>
</Button>

This will make the ContextMenu open where you right clicked your mouse (on the button). Which I think might be your desired location right?

--- EDIT --- In that case use this:

<Button Name="btnFoo" Content="Foo" ContextMenuOpening="ContextMenu_ContextMenuOpening">
    <Button.ContextMenu>
        <ContextMenu Placement="Bottom">
            <MenuItem Header="Bar" />
        </ContextMenu>
    </Button.ContextMenu>
</Button>

And in code behind:

private void ContextMenu_ContextMenuOpening(object sender, ContextMenuEventArgs e)
{
    // Get the button and check for nulls
    Button button = sender as Button;
    if (button == null || button.ContextMenu == null)
        return;
    // Set the placement target of the ContextMenu to the button
    button.ContextMenu.PlacementTarget = button;
    // Open the ContextMenu
    button.ContextMenu.IsOpen = true;
    e.Handled = true;
}

You can reuse the method for multiple buttons and ContextMenu's..

Answer 3

You could use a <Menu />, styled as a Button and avoid the hassle with the ContextMenuService.