MyTetra Share
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Контекстное меню
Время создания: 11.01.2020 20:37
Текстовые метки: Контекстное меню, wpf
Раздел: Компьютер - C# - WPF
Запись: Kozlov-AE/Tetra/master/base/15787642796093ngwmqr/text.html на raw.githubusercontent.com

Context Menus in WPF

Context Menus can be defined on any WPF controls by setting the ContextMenu property to an instance of a ContextMenu. The items of a context menu are normal MenuItems.

 

<RichTextBox>

<RichTextBox.ContextMenu>

<ContextMenu>

<MenuItem Command="Cut">

<MenuItem.Icon>

<Image Source="Images/cut.png" />

</MenuItem.Icon>

</MenuItem>

<MenuItem Command="Copy">

<MenuItem.Icon>

<Image Source="Images/copy.png" />

</MenuItem.Icon>

</MenuItem>

<MenuItem Command="Paste">

<MenuItem.Icon>

<Image Source="Images/paste.png" />

</MenuItem.Icon>

</MenuItem>

</ContextMenu>

</RichTextBox.ContextMenu>

</RichTextBox>

 

 

Show ContextMenus on a disabled controls

If you rightclick on a disabled control, no context menu is shown by default. To enable the context menu for disabled controls you can set the ShowOnDisabled attached property of the ContextMenuService to True.

 

<RichTextBox IsEnabled="False" ContextMenuService.ShowOnDisabled="True">

<RichTextBox.ContextMenu>

<ContextMenu>

...

</ContextMenu>

</RichTextBox.ContextMenu>

</RichTextBox>

 

 

Merge ContextMenus

If you want to fill a menu with items coming from multiple sources, you can use the CompositeCollection to merge multiple collection into one.

 

<Window xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"

xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"

xmlns:sys="clr-namespace:System;assembly=mscorlib">

<Grid Background="Transparent">

<Grid.Resources>

<x:Array Type="{x:Type sys:Object}" x:Key="extensions">

<Separator />

<MenuItem Header="Extension MenuItem 1" />

<MenuItem Header="Extension MenuItem 2" />

<MenuItem Header="Extension MenuItem 3" />

</x:Array>

</Grid.Resources>

<Grid.ContextMenu>

<ContextMenu>

<ContextMenu.ItemsSource>

<CompositeCollection>

<MenuItem Header="Standard MenuItem 1" />

<MenuItem Header="Standard MenuItem 2" />

<MenuItem Header="Standard MenuItem 3" />

<CollectionContainer Collection="{StaticResource extensions}" />

</CompositeCollection>

</ContextMenu.ItemsSource>

</ContextMenu>

</Grid.ContextMenu>

</Grid>

</Window>

 

 

How to bind a Command on a ContextMenu within a DataTemplate using MVVM

Since the Popuup control has it's separate visual tree, you cannot use find ancestor to find the Grid. The trick here is to use the PlacementTarget property, that contains the element, the ContextMenu is aligned to, what is the Grid in our case.

But this is only half of the solution. Because of the data template, the DataContext is set to a dataitem, and not the view model. So you need another relative source lookup, to find the view model. Trick Nr. 2 is to use the Tag property to bind the view model from outside to the grid, which is the PlacementTarget used above. And there we are.

 

<DataTemplate>

<Grid Tag="{Binding DataContext, RelativeSource={RelativeSource AncestorType={x:Type ListBox}}}">

<Grid.ContextMenu>

<ContextMenu DataContext="{Binding Path=PlacementTarget.Tag, RelativeSource={RelativeSource Self}}">

<MenuItem Content="Cut" Command="{Binding CutCommand}" />

<MenuItem Content="Copy" Command="{Binding CopyCommand}" />

<MenuItem Content="Paste" Command="{Binding PasteCommand}" />

</ContextMenu>

</Grid.ContextMenu>

</Grid>

</DataTemplate>

 

 

How to open a context menu from code

The following sample shows you how to open a context menu of a control programmatically:

 

private void OpenContextMenu(FrameworkElement element)

{

if( element.ContextMenu != null )

{

element.ContextMenu.PlacementTarget = element;

element.ContextMenu.IsOpen = true;

}

}

 

 
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