如何通过右键单击更改图像的可见性 [英] how to change visibility of Image with right click

问题描述

我有两个图片.
我的XAMl代码:

I have two Images.
My XAMl code:

<Image Height="13" HorizontalAlignment="Left" Margin="284,236,0,0" Name="CBoff2"  Stretch="Fill" VerticalAlignment="Top" Width="18" Source="/ConfigurationWindowsDemo;component/Images/SymbolsOFF1.jpg"/>
<Image Height="13" HorizontalAlignment="Left" Margin="284,236,0,0" Name="CB2"  Stretch="Fill" VerticalAlignment="Top" Width="18"  Source="/ConfigurationWindowsDemo;component/Images/SymbolsON.jpg" />

我要的是当我们右键单击"CB2"时，"CBoff2"应该可见，反之亦然.

What I Want is when we right click to "CB2", "CBoff2" should be visible and vice Versa. How can we do that?

推荐答案

<Image MouseRightButtonDown="image1_MouseRightButtonDown" Source="Sunset.jpg" Grid.Row="2" Margin="156,141,0,55" Name="image1" Stretch="Fill" HorizontalAlignment="Left" Width="139.83" />

该处理程序将禁用image2.

The handler will disable the image2.

private void image1_MouseRightButtonDown(object sender, MouseButtonEventArgs e)
{
    image2.Visibility = Visibility.Hidden;
}

但是，当另一个图像被隐藏时，您将如何做呢?

But how are you going to do vice-versa when the other image gets hidden.

我在图像控件中看不到任何事件处理程序.
您需要添加一个.

I don''t see any event handler in your image control.
You need to add one.

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