How can you set IsHitTestVisible to True on a child of something where it's set to false?

问题

I have a message which is within a border and is displayed in front of a MDI client area. Because it's just a message, I set it's IsHitTestVisible to false. For convenience, I've also included a button within it so the user can simply click the button to start a new diagram.

However, because the border has its IsHitTestVisible set to False, it shorts out the True value of the button so the user can't click on the button.

That said, how can you make a control invisible to the mouse, while still allowing one of its children to receive mouse events?

Here's a screenshot:

We want the blue area with text to be hit-test invisible, but we want the button to still be able to be pressed. Thoughts? (And yes, I already know about stacking the two items in a grid, but that messes with the layout.

回答1:

Ok, doesn't look like the question can be solved as asked, and therefore it looks like we have to revert to the layered method. In that case, to compensate for text changes, you have to set the vertical alignment of the button to Bottom, and ensure there's extra padding on the bottom edge of the border or bottom margin on the TextBlock) to ensure the label always floats above the button. You manage the side margins/padding for both using the same principles.

This will give us what we're after, but just seems pretty cumbersome. Again, I wish there was some way to specify IsHitTestVisible (or IsEnabled for that matter) with a 'Force' somehow that re-establishes them (and below) as the owner of the behaviors. A person can dream. Until then... layers it is!

回答2:

AFAIK you can't. you can do it however with IsEnabledProperty, and still with a little workaround. you have to create your own button like this:

Edited 9/1/2013 :

 public partial class Window1 : Window
{
    public Window1()
    {
        InitializeComponent();
        DataContext = this;
    }

    private void ButtonBase_OnClick(object sender, RoutedEventArgs e)
    {
        MessageBox.Show("hello");
    }
}

public class MyButton : Button
{
    static MyButton()
    {
        IsEnabledProperty.OverrideMetadata(typeof(MyButton),
        new UIPropertyMetadata(true,(o, args) => { },(o, value) => value));
    }
}

XAML:

<BorderIsEnabled="False">
    <wpfApplication2:MyButton Margin="61,95,88,100" Click="ButtonBase_OnClick">Open a new diagram</wpfApplication2:MyButton>
</Border>

回答3:

DISCLAIMER: It looks it wont solve exact problem that OP has but i will keep it here because it might be helpful for others with similar issue.

Clicking on grid should not register any mouse events on grid.

<Grid Background="{x:Null}">
   <Button/>
<Grid/>

EDIT: When you want background to be visibile then i would suggest this: For demostration i set cursor="hand" on border

<Canvas>
    <Border Height="300" Width="300" Background="Red" Cursor="Hand" IsHitTestVisible="False"/>
    <Button Content="click"/>
</Canvas>

<Grid>
    <Border  Background="Red" Cursor="Hand" IsHitTestVisible="False" />
    <Button Content="click" HorizontalAlignment="Center"/>
</Grid>

It both cases the trick is that the control (button) is not child of parent with IsHitTestVisible="False" and it is just overlaping the background control

来源：https://stackoverflow.com/questions/12148286/how-can-you-set-ishittestvisible-to-true-on-a-child-of-something-where-its-set

标签

wpf

layout

hittest