C# 如何始终在WPF中的切换按钮下放置弹出窗口
当我点击按钮时,我想在切换按钮下放置一个弹出窗口。在这个弹出窗口中,我想添加按钮和其他控件。但是,当我调整主窗口的大小时,如何确保弹出窗口始终位于我的切换按钮下呢 我的XAML代码:C# 如何始终在WPF中的切换按钮下放置弹出窗口,c#,wpf,popup,togglebutton,C#,Wpf,Popup,Togglebutton,当我点击按钮时,我想在切换按钮下放置一个弹出窗口。在这个弹出窗口中,我想添加按钮和其他控件。但是,当我调整主窗口的大小时,如何确保弹出窗口始终位于我的切换按钮下呢 我的XAML代码: <ToggleButton x:Name="userBtn" Margin="610,25,378,0" VerticalAlignment="Top" Height="29" Grid.Column="2" >
<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>
使用
Placement
和PlacementTarget
弹出窗口的属性
这里需要注意的是,如果在弹出窗口
打开时移动窗口,它将不会移动。。您可以将IsOpen
属性绑定到ToggleButton
的IsChecked
并将StaysOpen
=False。这样,弹出窗口将自动关闭,将按钮向后切换:),如果要执行此操作,您也不需要C代码:)
或者这里是另一个SO帖子的链接,如果它应该保持开放,无论情况如何:
希望有帮助:)谢谢您的回答@faztp12。弹出窗口在正确的位置,但当我移动主窗口时,它不会遵循:(完美解决方案:)非常感谢。但是如果没有C#code,它将无法工作?不再需要C#code,因为
弹出窗口
绑定到切换按钮
s IsChecked属性:)删除UserBtn#u单击事件;)我删除了UserBtn_Click事件,但它不起作用?:(您是否在弹出窗口中添加了'IsOpen=“{Binding IsChecked,ElementName=Userbtn}'”
?:)
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;
}
<Popup Placement="Bottom" PlacementTarget="{Binding ElementName=myToggleButton}" IsOpen="{Binding IsChecked, ElementName=Userbtn}" StaysOpen="False" />