Enable command binding for TextBlock
Asked Answered
A

3

15

I am developing a WPF application, and have a TextBlock which I want to use command binding to trigger a command on when clicked. What's the best way to achieve this?

  • The TextBlock-control does not have a Command property, but it does have a CommandManager. What is this? Can it be used for command bindings? I've seen many other controls as well with this property..

  • Is there some control I have overseen that can be used? Is it e.g. recommended to use a button and style it to not look like a button?

  • Is there some controls supporting Command bindings which I can wrap around the TextBlock?

  • Should I create a custom control which basically is a TextBlock, but with extra properties Command and CommandArgument which enables command binding on e.g. the MouseLeftButtonDown property.

Achernar answered 24/11, 2009 at 13:12 Comment(0)
G
17

Is there some control I have overseen that can be used? Is it e.g. recommended to use a button and style it to not look like a button?

Yes. The simplest approach would be to re-template a button to act like a TextBlock and leverage the command property on the button class.

Something like this:

<ControlTemplate TargetType="Button">
        <TextBlock Text="{TemplateBinding Content}" />
    </ControlTemplate>
...
<Button Content="Foo" Command="{Binding Bar}" />
Guttersnipe answered 24/11, 2009 at 16:57 Comment(2)
Thanks. Works like expected - and it's actually pretty neat..! I've read up on ControlTemplates in "WPF Unleased" now, and this feels right. What does the "Text="{TemplateBinding Content}" " do anyway? Bind what I set to the buttons Content property to the Text property on the TextBlock?Achernar
Yup, that is exactly what it does.Guttersnipe
B
4

The below XAML can be used to add a Command Binding to a WPF TextBlock which will then work on a mouse action.

<TextBlock FontWeight="Bold" Text="Header" Cursor="Hand">
    <TextBlock.InputBindings>
        <MouseBinding Command="ApplicationCommands.Cut" MouseAction="LeftClick"/>
    </TextBlock.InputBindings>
</TextBlock>

The Command can be one of the built-in application commands, which would use the syntax as shown above, or it can be a custom Command that inherits from the ICommand Interface. In that case, the syntax would be:

<MouseBinding Command="{Binding myCustomCommand}" MouseAction="LeftClick"/>

The MouseAction doesn't provide any Intellisense tips on what to put there (in VS2015), so you have to do a little digging to get the valid enumerations.

As of .NET 4.5, the valid entries for MouseAction are:

  • LeftClick - left mouse button click.
  • LeftDoubleClick - left mouse button double-click.
  • MiddleClick - middle mouse button click.
  • MiddleDoubleClick - middle mouse button double-click.
  • None - No action.
  • RightClick - right mouse button click.
  • RightDoubleClick - right mouse button double-click.
  • WheelClick - A mouse wheel rotation.

Constants shown above are taken from MSDN.

Bendicty answered 9/2, 2016 at 18:13 Comment(3)
While this code may answer the question, it would be better to explain how it solves the problem without introducing others and why to use it. Code-only answers are not useful in the long run.Servo
Thanks for the suggestion @BenjaminW. This is my first answer in the community :-)Bendicty
@Praveen, good answer. It was exactly what I needed. I edited your answer to include more description and usage tips.Trichloroethylene
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<Window.Resources>
<CommandBinding x:Key="binding" Command="ApplicationCommands.Save" Executed="SaveCommand" CanExecute="SaveCommand_CanExecute" />
</Window.Resources>


<TextBox Margin="5" Grid.Row="2" TextWrapping="Wrap" AcceptsReturn="True" TextChanged="txt_TextChanged">
<TextBox.CommandBindings>
<StaticResource ResourceKey="binding"></StaticResource>
</TextBox.CommandBindings>
</TextBox>

Have you seen http://www.java2s.com/Tutorial/CSharp/0470__Windows-Presentation-Foundation/BindTextBoxsavecommandtoCommandBinding.htm

Atom answered 2/8, 2012 at 23:21 Comment(1)
Please explain how this answer addresses the questionHancock

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