<Button Name="btnFoo" Content="Foo" >
<Button.ContextMenu Placement="Bottom" PlacementTarget="btnFoo">
<MenuItem Header="Bar" />
</Button.ContextMenu>
</Button>
gives me a runtime error 'UIElement' type does not have a public TypeConverter class
I also tried
<Button Name="btnFoo" Content="Foo" >
<Button.ContextMenu Placement="Bottom" PlacementTarget="{Binding ElementName=btnFoo}">
<MenuItem Header="Bar" />
</Button.ContextMenu>
</Button>
and that put the ContextMenu in the top left corner of my screen, rather than at the Button
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..
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