How do get menu to open to the left in WPF?

Question

I have a menu (with menuitems) in WPF. Unfortunately when I click on the menu heading it opens the menu to the right. The problem is that there is stuff on the right that I don't want it to overlap. How do I tell WPF to open the menu to the left? Do I need to do a control template? (control templates seem so heavy handed for such basic style changes).

Thanks!

KSG

Answer 1

While you can create a ControlTemplate to do this like they do here, I agree that it is a cumbersome method just to modify one value on a part of the MenuItems. Instead, I think that this is a great place to use an AttachedProperty. We can create something just like the ContextMenuService, but for Popups (In fact, I'm somewhat surprised that it isn't built in).

To change where the popup is opening, we're going to want to set the Popup's PlacementMode. We can use the propa shortcut to generate our AttachedProperty(or properties if you want to implement the rest). We need to add a callback to our PropertyMetadata, but if the AttachedProperty is set inline on the control in XAML then the callback will fire before the whole control is fully constructed. To ensure the MenuItem's template is applied, and the Popup exists before we try and set it's value, we can just attach to the Loaded event if it isn't already loaded. Once it is loaded, we want to retrieve the Popup from the template, and if we look at the MenuItem class we can see that it has a TemplatePartAttribute defining the Popup's name as "PART_Popup". Once we have that, we can set the PlacementMode on the MenuItem's Popup.

    public static PlacementMode GetMenuPlacement(DependencyObject obj)
 {
  return (PlacementMode)obj.GetValue(MenuPlacementProperty);
 }

 public static void SetMenuPlacement(DependencyObject obj, PlacementMode value)
 {
  obj.SetValue(MenuPlacementProperty, value);
 }

 // Using a DependencyProperty as the backing store for MenuPlacement.  This enables animation, styling, binding, etc...
 public static readonly DependencyProperty MenuPlacementProperty =
  DependencyProperty.RegisterAttached("MenuPlacement",
  typeof(PlacementMode),
  typeof(Window1),
  new FrameworkPropertyMetadata(PlacementMode.Bottom, FrameworkPropertyMetadataOptions.Inherits, new PropertyChangedCallback(OnMenuPlacementChanged)));

 private static void OnMenuPlacementChanged(DependencyObject o, DependencyPropertyChangedEventArgs e)
 {
  var menuItem = o as MenuItem;
  if (menuItem != null)
  {
   if (menuItem.IsLoaded)
   {
    SetPopupPlacement(menuItem, (PlacementMode)e.NewValue);
   }
   else
   {
    menuItem.Loaded += new RoutedEventHandler((m, v) => SetPopupPlacement(menuItem, (PlacementMode)e.NewValue));
   }
  }
 }

 private static void SetPopupPlacement(MenuItem menuItem, PlacementMode placementMode)
 {
  Popup popup = menuItem.Template.FindName("PART_Popup", menuItem) as Popup;
  if (popup != null)
  {
   popup.Placement = placementMode;
  }
 }

Now that we have our AttachedProperty, it's easy to change the Popup placement in the UI.

<Menu>
 <MenuItem Header="Item 1"
     local:Window1.MenuPlacement="Right">
  <MenuItem Header="SubItem 1" />
  <MenuItem Header="SubItem 2" />
  <MenuItem Header="SubItem 3" />
  <MenuItem Header="SubItem 4" />
 </MenuItem>
 <MenuItem Header="Item 2"
     local:Window1.MenuPlacement="Left">
  <MenuItem Header="SubItem 5" />
  <MenuItem Header="SubItem 6" />
  <MenuItem Header="SubItem 7" />
  <MenuItem Header="SubItem 8" />
 </MenuItem>
 <MenuItem Header="Item 3"
     local:Window1.MenuPlacement="Mouse">
  <MenuItem Header="SubItem 9" />
  <MenuItem Header="SubItem 10" />
  <MenuItem Header="SubItem 11" />
  <MenuItem Header="SubItem 12" />
 </MenuItem>
</Menu>