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Home/ Questions/Q 641669
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T21:05:17+00:00 2026-05-13T21:05:17+00:00

I’m using MVVM to bind views to objects in a Tree. I have a

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I’m using MVVM to bind views to objects in a Tree. I have a base class that implements the items in my tree, and that base class has a ContextMenu property:

    public IEnumerable<IMenuItem> ContextMenu
    {
        get
        {
            return m_ContextMenu;
        }
        protected set
        {
            if (m_ContextMenu != value)
            {
                m_ContextMenu = value;
                NotifyPropertyChanged(m_ContextMenuArgs);
            }
        }
    }
    private IEnumerable<IMenuItem> m_ContextMenu = null;
    static readonly PropertyChangedEventArgs m_ContextMenuArgs =
        NotifyPropertyChangedHelper.CreateArgs<AbstractSolutionItem>(o => o.ContextMenu);

The View that binds to the base class (and all derived classes) implements a ContextMenu that binds to that property:

<ContextMenu x:Name="contextMenu" ItemsSource="{Binding Path=(local:AbstractSolutionItem.ContextMenu)}"
             IsEnabled="{Binding Path=(local:AbstractSolutionItem.ContextMenuEnabled)}"
             ItemContainerStyle="{StaticResource contextMenuStyle}"/>

Each item in the menu is bound to an IMenuItem object (a ViewModel for the menu items). When you click on the menu item, it uses Commands to execute a command on the base object. This all works great.

However, once the command is executing on the IMenuItem class, it sometimes needs to get a reference to the object that the user right clicked on to bring up the context menu (or the ViewModel of that object, at least). That’s the whole point of a context menu. How should I go about passing the reference of the tree item ViewModel to the MenuItem ViewModel? Note that some context menus are shared by many objects in the tree.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T21:05:18+00:00Added an answer on May 13, 2026 at 9:05 pm

    I solved this by handling the ContextMenuOpening event on the parent control (the one that owned the ContextMenu in the View). I also added a Context property to IMenuItem. The handler looks like this:

        private void stackPanel_ContextMenuOpening(
            object sender, ContextMenuEventArgs e)
        {
            StackPanel sp = sender as StackPanel;
            if (sp != null)
            {
                // solutionItem is the "context"
                ISolutionItem solutionItem =
                    sp.DataContext as ISolutionItem;
                if (solutionItem != null) 
                {
                    IEnumerable<IMenuItem> items = 
                        solutionItem.ContextMenu as IEnumerable<IMenuItem>;
                    if (items != null)
                    {
                        foreach (IMenuItem item in items)
                        {
                            // will automatically set all 
                            // child menu items' context as well
                            item.Context = solutionItem;
                        }
                    }
                    else
                    {
                        e.Handled = true;
                    }
                }
                else
                {
                    e.Handled = true;
                }
            }
            else
            {
                e.Handled = true;
            }
        }
    

    This takes advantage of the fact that there can only be one ContextMenu open at a time.

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