wpf - 在 WPF TreeView 中更改数据的上下文菜单

Question

有没有办法在 a 中指定根据项目绑定到的数据上的属性TreeView使用HierarchicalDataTemplate不同的值？ContextMenu

例如，如果为真则显示一个，ContextMenu如果为真则显示另一个，等等。Item.IsFileItem.IsFolder

Answer 1

这是 ListBox 的示例，我认为您可以轻松修改它以使用 TreeView。

XAML：

...

<Window.Resources>
    <ContextMenu x:Key="FileContextMenu">
        ...
    </ContextMenu>
    <ContextMenu x:Key="DirContextMenu">
        ...
    </ContextMenu>

    <local:ItemToContextMenuConverter x:Key="ContextMenuConverter" />        
</Window.Resources>

...

<ListBox x:Name="SomeList">
    <ListBox.ItemTemplate>
        <DataTemplate>                          
            <Label Content="{Binding Path=Name}" ContextMenu="{Binding Converter={StaticResource ContextMenuConverter}}"/>
        </DataTemplate>
    </ListBox.ItemTemplate>
</ListBox>

代码：

class Item
{
    public string Name { get; set; }
    public bool IsFile { get; set; }
}

[ValueConversion(typeof(Item), typeof(ContextMenu))]
public class ItemToContextMenuConverter : IValueConverter
{
    public static ContextMenu FileContextMenu;
    public static ContextMenu DirContextMenu;

    public object Convert(object value, Type targetType, object parameter, CultureInfo culture)
    {
        Item item = value as Item;
        if (item == null) return null;

        return item.IsFile ? FileContextMenu : DirContextMenu;
    }

    public object ConvertBack(object value, Type targetType, object parameter, CultureInfo culture)
    {
        throw new Exception("The method or operation is not implemented.");
    }
}

private void Window_Loaded(object sender, RoutedEventArgs e)
    {
        ItemToContextMenuConverter.FileContextMenu 
            = this.Resources["FileContextMenu"] as ContextMenu;
        ItemToContextMenuConverter.DirContextMenu 
            = this.Resources["DirContextMenu"] as ContextMenu;

        List<Item> items = new List<Item>();
        items.Add(new Item() { Name = "First", IsFile = true });
        items.Add(new Item() { Name = "Second", IsFile = false });

        SomeList.ItemsSource = items;
    }

Answer 2

嗨，我在 TreeView 上做类似的事情，我不喜欢在每个项目上执行 ItemToContextMenuConverter，即使它没有被使用。在一个小项目中可能没问题，但如果为每个 MenuItem 添加启用/禁用代码，它可能会很慢。

这可能不是最好的（我刚开始使用 WPF），但我会与您分享。

菜单资源：

<Window.Resources>
  <ContextMenu x:Key="MnuFolderFavorites" StaysOpen="True">
    <MenuItem Header="Remove from Favorites" Click="MnuFolder_RemoveFromFavorites_Click"></MenuItem>
  </ContextMenu>
  <ContextMenu x:Key="MnuFolder" StaysOpen="True">
    <MenuItem Header="New Folder"></MenuItem>
    <MenuItem Header="Rename" x:Name="MnuFolderRename" Click="MnuFolder_Rename_Click"></MenuItem>
    <MenuItem Header="Add to Favorites" Click="MnuFolder_AddToFavorites_Click"></MenuItem>
  </ContextMenu>
</Window.Resources>

树视图：

<TreeView x:Name="TvFolders">
  <TreeView.ItemTemplate>
    <HierarchicalDataTemplate DataType="{x:Type data:Folder}" ItemsSource="{Binding Items}">
      <StackPanel Orientation="Horizontal" PreviewMouseRightButtonDown="TvFoldersStackPanel_PreviewMouseRightButtonDown">
        <Image Width="20" Height="20" Source="{Binding ImagePath}" />
        <TextBlock Text="{Binding Title}" Margin="5,0,0,0" />
      </StackPanel>
    </HierarchicalDataTemplate>
  </TreeView.ItemTemplate>
</TreeView>

代码：

private void TvFoldersStackPanel_PreviewMouseRightButtonDown(object sender, MouseButtonEventArgs e) {
  ((StackPanel) sender).ContextMenu = null;
  Data.Folder item = (Data.Folder) ((StackPanel) sender).DataContext;
  if (!item.Accessible) return;
  if (item.Parent != null && item.Parent.Title.Equals("Favorites")) {
    ((StackPanel) sender).ContextMenu = MainWindow.Resources["MnuFolderFavorites"] as ContextMenu;
  } else {
    ((StackPanel) sender).ContextMenu = MainWindow.Resources["MnuFolder"] as ContextMenu;
    foreach (MenuItem menuItem in ((StackPanel) sender).ContextMenu.Items) {
      switch (menuItem.Name) {
        case "MnuFolderRename": {
          menuItem.IsEnabled = item.Parent != null;
          break;
        }
      }
    }
  }
}

private void MnuFolder_RemoveFromFavorites_Click(object sender, RoutedEventArgs e) {
  string path = ((Data.Folder)((MenuItem)sender).DataContext).FullPath;
  Settings.Default.FolderFavorites.Remove(path);
  Settings.Default.Save();
  FavoritesFolder?.AddFavorites(true);
}