WPF Context Menus in Caliburn Micro

Question

I\'m trying to get a context menu within a ListBox ItemTemplate to call a method on the parent view model, passing in the item that was clicked on as a parameter. I have thi

Answer 1

Adding to Jason's answer, if you're going to be using the same data context as the control, then you can just bind the DataContext instead of a Tag

<Grid>
    <Grid.ContextMenu>
        <ContextMenu cal:Action.TargetWithoutContext="{Binding Path=PlacementTarget.DataContext, RelativeSource={RelativeSource Self}}">
            <MenuItem Header="Open" 
                      cal:Message.Attach="Open($source)">
            </MenuItem>
        </ContextMenu>
    </Grid.ContextMenu>
</Grid>

$source The actual FrameworkElement that triggered the ActionMessage to be

You can see more info about the $source convention in here: https://caliburnmicro.com/documentation/cheat-sheet

Answer 2

Using the information I found on the Caliburn Micro site I modified your XAML to look like this:

  <Grid Background="White" HorizontalAlignment="Stretch" Height="200" Name="GridLayout">       
    <ListBox x:Name="ListBoxItems">            
        <ListBox.ItemTemplate>
            <DataTemplate>
                <Grid Tag="{Binding DataContext, ElementName=GridLayout}">
                    <Grid.ContextMenu>
                        <ContextMenu Name="cm" cal:Action.TargetWithoutContext="{Binding Path=PlacementTarget.Tag, RelativeSource={RelativeSource Self}}">
                            <MenuItem Header="Open" 
                              cal:Message.Attach="Open($dataContext)">
                            </MenuItem>
                        </ContextMenu>
                    </Grid.ContextMenu>

                    <TextBlock VerticalAlignment="Center" >
                .. text..
                    </TextBlock>
                </Grid>
            </DataTemplate>
        </ListBox.ItemTemplate>
    </ListBox>
</Grid>

And my view model:

    public List<string> ListBoxItems { get; set; }
    public ShellViewModel()
    {
        ListBoxItems = new List<string> {"One", "Two", "Three"};          
    }

    public void Open(object source)
    {
        MessageBox.Show((string) source);
    }

Open was successfully called with the appropriate strings from the list box.