WPF弹出窗口中的子MenuItem

Question

我在XAML的弹出窗口中创建了MenuItems。该菜单项包含SubMenu项。单击切换按钮时弹出窗口。我无法在弹出窗口中打开MenuItem的子MenuItem。

$

<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>

请帮我解决这个问题。

Answer 1

将MenuItem包裹在Menu中，它会起作用。