我正在尝试在右键单击元素的顶部打开上下文菜单。但它只在鼠标点击位置打开。
<TextBlock Grid.Row="1" Grid.Column="0" VerticalAlignment="Center" x:Name="testeTB" Text="Name:">
<TextBlock.ContextMenu>
<ContextMenu>
<ContextMenu.Style>
<Style TargetType="ContextMenu">
<Setter Property="Placement" Value="Top"></Setter>
<Setter Property="PlacementTarget" Value="{Binding ElementName=testeTB}"></Setter>
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="ContextMenu">
<controls:PopupMenu>
<controls:PopupMenu.Children>
<wpf:ArbeitListBoxItem Content="Add Messages"></wpf:ArbeitListBoxItem>
<wpf:ArbeitListBoxItem Content="Remove Messages"></wpf:ArbeitListBoxItem>
</controls:PopupMenu.Children>
</controls:PopupMenu>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>
</ContextMenu.Style>
</ContextMenu>
</TextBlock.ContextMenu>
</TextBlock>
正如您在我的示例中所看到的,我尝试将Placement设置为&#34; Top&#34;和 还使用textblock元素名称设置PlacementTarget,但它 没有工作。
是否可以使用ContextMenu来完成?或者我应该使用弹出窗口?我更喜欢使用ContextMenu
答案 0 :(得分:1)
尝试设置目标Control的ContextMenuService.Placement
属性,这是一个XAML示例:
<Grid>
<TextBlock
Padding="10 5"
Background="Lime"
ContextMenuService.Placement="Right"
Grid.Row="1"
Grid.Column="0"
HorizontalAlignment="Center"
VerticalAlignment="Center"
Text="Name:">
<TextBlock.ContextMenu>
<ContextMenu>
<ContextMenu.Template>
<ControlTemplate>
<Border Background="Red">
<TextBlock Text="Context menu content..." Margin="20 15" />
</Border>
</ControlTemplate>
</ContextMenu.Template>
</ContextMenu>
</TextBlock.ContextMenu>
</TextBlock>
</Grid>