WPF 弹出窗口中的子菜单项

Question

我在 XAML 的弹出窗口内创建了菜单项。该菜单项包含子菜单项。弹出窗口在单击切换按钮时打开。我无法在弹出窗口中打开菜单项的子菜单项。

$

<ToggleButton Name="button"
                  Width="30"
                  Height="30"
                  Click="Button_Click">
        <Image Width="30" Height="30">
            <Image.Source>
                <BitmapImage UriSource="/WpfApplication4;component/Images/Filter.png" />
            </Image.Source>
        </Image>
    </ToggleButton>
    <Popup Name="PART_Popup"
           Placement="Bottom"
           PlacementTarget="{Binding ElementName=button}">
        <Border Background="White"
                BorderBrush="Gray"
                BorderThickness="1">
            <StackPanel Orientation="Vertical">

                <MenuItem  Width="270"
                          Margin="5"
                          Header="ClearFilter">
                        <MenuItem.Icon>
                            <Image>
                                <Image.Source>
                                    <BitmapImage UriSource="/WpfApplication4;component/Images/ClearFilter.png" />
                                </Image.Source>
                            </Image>
                        </MenuItem.Icon>
                    <MenuItem  Header="SubMenu" />
                        <MenuItem Header="SubMenu1" />
                    </MenuItem>
                               </StackPanel>
        </Border>
    </Popup>

请帮助我解决这个问题。

Answer 1

将

MenuItem包装在Menu中，它将起作用。