显示ContextMenu

Question

尽管找了一段时间，我还是很难找到好的文档。

我希望在我的应用程序中有一个上下文菜单，可以复制其他点击按住上下文菜单的行为，比如将应用程序固定在应用程序列表的开始屏幕上。

这是我的上下文菜单：

                <toolkit:ContextMenuService.ContextMenu>
                    <toolkit:ContextMenu x:Name="sectionContextMenu">
                        <toolkit:MenuItem Header="Hide this section from this list" />
                    </toolkit:ContextMenu>
                </toolkit:ContextMenuService.ContextMenu>

我如何让它显示出来呢？

Answer 1

上下文菜单需要附加到您希望用户点击并按住的元素。

<Border Margin="0,12" BorderBrush="{StaticResource PhoneForegroundBrush}" BorderThickness="2" Background="Transparent" VerticalAlignment="Center" Padding="16">
   <toolkit:ContextMenuService.ContextMenu>
      <toolkit:ContextMenu x:Name="sectionContextMenu">
         <toolkit:MenuItem Header="Hide this section from this list" />
      </toolkit:ContextMenu>
   </toolkit:ContextMenuService.ContextMenu>
   <TextBlock Text="Tap and hold here to invoke a ContextMenu" Style="{StaticResource PhoneTextNormalStyle}"/>
</Border>

用户现在可以通过轻触并按住此Border元素的内容来调用上下文菜单。

Answer 2

根据内容为不同项目提供独特的上下文菜单。

private ContextMenu CreateContextMenu(ListBoxItem lbi)
{
    ContextMenu contextMenu = new ContextMenu();
    ContextMenuService.SetContextMenu(lbi, contextMenu);
    contextMenu.Padding = new Thickness(0);

    string item_1 = "item 1";
    if(lbi.Content is string) {
        item_1 = lbi.Content as string;
    }
    contextMenu.ItemsSource = new List<string> { item_1, "item 2", "item 3" };
    contextMenu.IsOpen = true;
    return contextMenu;
}

private void Results_SelectionChanged(object sender, SelectionChangedEventArgs e)
{
    if (Results.SelectedIndex == -1) return;
    int index = Results.SelectedIndex;

    ListBoxItem lbi = Results.ItemContainerGenerator.ContainerFromIndex(index) as ListBoxItem;

    CreateContextMenu(lbi);
    Results.SelectedIndex = -1;
}