如何始终将 PopUp 放在 WPF 中的 ToggleButton 下

Question

我想在单击切换按钮时在其下放置一个弹出窗口。在此弹出窗口中，我想添加按钮和其他控件。但是，当我调整 mian window 大小时，如何确保弹出窗口 window 始终位于我的切换按钮下方。

我的XAML代码：

     <ToggleButton
    x:Name="userBtn"
    Margin="610,25,378,0"
    VerticalAlignment="Top"            
    Height="29"
    Grid.Column="2" 
    >
        <ToggleButton.Resources>
            <BitmapImage x:Key="imgNormal" UriSource="/Resources/home.jpg"/>
            <BitmapImage x:Key="imgHover" UriSource="/Resources/home_checked.jpg"/>
            <BitmapImage x:Key="imgChecked" UriSource="/Resources/home_checked.jpg"/>
        </ToggleButton.Resources>
        <ToggleButton.Style>
            <Style TargetType="ToggleButton">
                <Setter Property="Template">
                    <Setter.Value>
                        <ControlTemplate TargetType="ToggleButton" >
                            <StackPanel Orientation="Horizontal" VerticalAlignment="Top" Width="150" Height="32">
                                <Image x:Name="PART_Image" Height="22" Width="22" RenderOptions.BitmapScalingMode="HighQuality" HorizontalAlignment="Center" Margin="0,0,0,0" Source="{StaticResource imgNormal}"/>
                                <TextBlock x:Name="PART_TEXT" FontFamily="Arial" Foreground="#FFAFADAD" FontSize="13" HorizontalAlignment="Center" Text="{x:Static properties:Resources.BtnDashboard}"  Margin="10,9,0,0"></TextBlock>
                            </StackPanel>
                            <ControlTemplate.Triggers>
                                <Trigger Property="IsChecked" Value="true">
                                    <Setter TargetName="PART_Image" Property="Source" Value="{StaticResource imgChecked}"/>
                                    <Setter TargetName="PART_TEXT" Property="Foreground" Value="#79B539" />
                                </Trigger>
                                <Trigger Property="IsMouseOver" Value="true">
                                    <Setter TargetName="PART_Image" Property="Source" Value="{StaticResource imgHover}"/>
                                    <Setter TargetName="PART_TEXT" Property="Foreground" Value="#79B539" />
                                </Trigger>
                                <Trigger Property="IsEnabled" Value="false">
                                    <Setter TargetName="PART_Image" Property="Opacity" Value="0.6"/>
                                </Trigger>
                            </ControlTemplate.Triggers>
                        </ControlTemplate>
                    </Setter.Value>
                </Setter>
            </Style>
        </ToggleButton.Style>
    </ToggleButton>

    <Popup Name="UserMenuPopUp" PopupAnimation="Fade" Width="280" Height="auto" Margin="20" AllowsTransparency="True" HorizontalAlignment="Left" Placement="bottom" PlacementTarget="{Binding ElementName=userBtn}" IsOpen="{Binding IsChecked, ElementName=Userbtn}" StaysOpen="false" >
        <Border Margin="2,0,48,0" BorderThickness="1" Background="#EEEEEE" Height="105">
            <Grid>

            </Grid>
        </Border>
    </Popup>

点击切换按钮的 C# 代码：

    private void UserBtn_Click(object sender, RoutedEventArgs e)
    {
        // Get location, adn sender
        var element = (System.Windows.Controls.Primitives.ToggleButton)sender;
        var location = element.PointToScreen(new Point(0, 0));

        // Set popup location
        UserMenuPopUp.HorizontalOffset = (location.X);
        UserMenuPopUp.VerticalOffset = (location.Y);
        UserMenuPopUp.IsOpen = true;
    }

Answer 1

使用 Popup 的 Placement 和 PlacementTarget 属性 :)

<Popup Placement="Bottom" PlacementTarget="{Binding ElementName=myToggleButton}" IsOpen="{Binding IsChecked, ElementName=Userbtn}" StaysOpen="False" />

这里需要注意的是，如果 window 在 Popup 打开时被移动，它不会移动。你可以做的是绑定 IsOpen 属性到 ToggleButton 的 IsChecked 并设置 StaysOpen=False。这样做，弹出窗口将自动关闭，将按钮切换回来:)，如果你要这样做，你也不需要 C# 代码:)

OR 这是一个 link 到另一个 SO post 如果它应该保持打开状态，无论如何： Move a WPF Popup

希望对您有所帮助:)

如何始终将 PopUp 放在 WPF 中的 ToggleButton 下

How to always place PopUp under a ToggleButton in WPF

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