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C# WPF:右键单击时显示面板_C#_Wpf_Xaml - Fatal编程技术网

C# WPF:右键单击时显示面板

C# WPF:右键单击时显示面板,c#,wpf,xaml,C#,Wpf,Xaml,我试图在用户右键单击控件时,在用户控件中的光标位置“显示”WPF ViewBox。现在,我有代码: <!-- XAML --> <Viewbox Width="100" Visibility="Collapsed" x:Name="actionBox"> <StackPanel Orientation="Horizontal" HorizontalAlignment="Stretch"> <Button>Item ▼<

我试图在用户右键单击控件时,在用户控件中的光标位置“显示”WPF ViewBox。现在,我有代码:

<!-- XAML -->
<Viewbox Width="100" Visibility="Collapsed" x:Name="actionBox">
    <StackPanel Orientation="Horizontal" HorizontalAlignment="Stretch">
        <Button>Item ▼</Button>
        <Button>Permute ▼</Button>
        <Button>Generate ▼</Button>
   </StackPanel>
</Viewbox>
事件确实被炒了,但似乎什么也没发生。(MouseRightButtonDown=“…”位于XAML文件的不同部分。)


如何在WPF中编写此内容?

请查看上下文菜单

<ContextMenu Name="cm" StaysOpen="true">

                    <MenuItem Header="Item ▼"/>

                    <MenuItem Header="Permute ▼"/>

                    <MenuItem Header="Generate ▼"/>

</ContextMenu>

您甚至可以将命令与菜单项绑定,也可以创建子菜单

有关更多信息:


谢谢。我一直在寻找一个稍微不同的用户界面,但这似乎是最简单的方式做到这一点!
<ContextMenu Name="cm" StaysOpen="true">

                    <MenuItem Header="Item ▼"/>

                    <MenuItem Header="Permute ▼"/>

                    <MenuItem Header="Generate ▼"/>

</ContextMenu>