WPF 编程打开ListViewItem的ContextMenu丢失DataContext问题

ContextMenu属于WPF的弹出层控件，不在主可视化树中，仅当通过右键点击触发打开时，WPF会自动将ContextMenu的PlacementTarget属性设置为被点击的ListViewItem，此时内部命令绑定可以正常继承ListViewItem的DataContext。你通过ToggleButton绑定手动设置IsOpen=true时，系统不会自动赋值PlacementTarget，导致ContextMenu找不到绑定源，命令失效。

直接在ContextMenu声明中显式绑定DataContext与PlacementTarget即可，无需修改其他逻辑，修改后的代码如下：

<ListView>
 <ListView.ItemContainerStyle>
     <Style TargetType="{x:Type ListViewItem}">
         <Setter Property="ContextMenu">
             <Setter.Value>
                 <!-- 新增PlacementTarget和DataContext绑定 -->
                 <ContextMenu 
                     PlacementTarget="{Binding RelativeSource={RelativeSource FindAncestor, AncestorType={x:Type ListViewItem}}}"
                     DataContext="{Binding PlacementTarget.DataContext, RelativeSource={RelativeSource Self}}">
                     <MenuItem
                         Header="Restore"
                         Command="{Binding RestoreCommand}" />
                     <MenuItem
                         Header="Delete"
                         Command="{Binding DeleteCommand}"/>
                 </ContextMenu>
             </Setter.Value>
         </Setter>
     </Style>
 </ListView.ItemContainerStyle>

 <ListView.View>
     <GridView>
         <!-- My columns here -->
         <GridViewColumn>
             <GridViewColumn.CellTemplate>
                 <DataTemplate>
                     <ToggleButton IsChecked="{Binding Path=ContextMenu.IsOpen, RelativeSource={RelativeSource FindAncestor, AncestorType={x:Type ListViewItem}}}"/>
                 </DataTemplate>
             </GridViewColumn.CellTemplate>
         </GridViewColumn>
     </GridView>
 </ListView.View>
</ListView>

如果需要控制ContextMenu的弹出位置，还可以额外给ContextMenu设置Placement属性，比如设为Bottom让菜单在ListViewItem下方弹出。

内容的提问来源于stack exchange，提问作者Gary Francis