Setting a WPF ContextMenu's PlacementTarget property in XAML?

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>

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..

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