UWP Async MenuFlyout openopened

Question

我需要设置我的 MenuFlyout 程式化的方式。然而，我注意到，在我的函数中添加了异步代码后，函数的 MenuFlyout 除非我第二次右击该项目，否则不会显示其项目。我怎样才能设置我的 MenuFlyout 的异步方式？

private async void MenuFlyout_Opened(object sender, object e)
{
    var flyout = sender as MenuFlyout;
    Music music = flyout.Target.DataContext as Music;
    if (await Helper.FileNotExist(music.Path))
    {
        if (Removable)
        {
            flyout.Items.Clear();
            flyout.Items.Add(MenuFlyoutHelper.GetRemovableMenuFlyoutItem(music, this));
        }
        else
        {
            Helper.ShowAddMusicResultNotification(music.Name);
        }
        return;
    }
    if (Removable) MenuFlyoutHelper.SetRemovableMusicMenu(sender, this);
    else MenuFlyoutHelper.SetMusicMenu(sender, this);
    if (AllowReorder)
    {
        var item = new MenuFlyoutItem()
        {
            Text = Helper.Localize("Move To Top"),
            Icon = new SymbolIcon(Symbol.Upload)
        };
        item.Click += (s, args) =>
        {
            MediaHelper.MoveMusic(music.Index, 0);
        };
        flyout.Items.Add(item);
    }
}

一种方法是避免在此类函数中使用异步代码，并检查文件是否在 Click 物品的事件。但我有太多的项目。我认为这是个坏主意。

这段代码可以正常工作。

private void OpenMusicMenuFlyout(object sender, object e)
{
    if (Removable) MenuFlyoutHelper.SetRemovableMusicMenu(sender, this);
    else MenuFlyoutHelper.SetMusicMenu(sender, this);
    if (AllowReorder)
    {
        var flyout = sender as MenuFlyout;
        var item = new MenuFlyoutItem()
        {
            Text = Helper.Localize("Move To Top"),
            Icon = new SymbolIcon(Symbol.Upload)
        };
        item.Click += (s, args) =>
        {
            Music music = (s as MenuFlyoutItem).DataContext as Music;
            MediaHelper.MoveMusic(music.Index, 0);
        };
        flyout.Items.Add(item);
    }
}

---

这段代码 Dispatcher.RunAsync:

private async void OpenMusicMenuFlyout(object sender, object e)
{
    var flyout = sender as MenuFlyout;
    Music music = flyout.Target.DataContext as Music;
    if (await Helper.FileNotExist(music.Path))
    {
        if (Removable)
        {
            flyout.Items.Clear();
            flyout.Items.Add(MenuFlyoutHelper.GetRemovableMenuFlyoutItem(music, this));
        }
        else
        {
            Helper.ShowAddMusicResultNotification(music.Name);
        }
        return;
    }
    await Dispatcher.RunAsync(Windows.UI.Core.CoreDispatcherPriority.High, () =>
    {

        if (Removable) MenuFlyoutHelper.SetRemovableMusicMenu(sender, this);
        else MenuFlyoutHelper.SetMusicMenu(sender, this);
        if (AllowReorder)
        {
            var item = new MenuFlyoutItem()
            {
                Text = Helper.Localize("Move To Top"),
                Icon = new SymbolIcon(Symbol.Upload)
            };
            item.Click += (s, args) =>
            {
                MediaHelper.MoveMusic(music.Index, 0);
            };
            flyout.Items.Add(item);
        }
    });
}

Answer 1

很好的问题，为了测试，如果我们在上面的函数中调用异步函数 OpenMusicMenuFlyout 事件处理程序。 MenuFlyout 添加操作会被中断，UI渲染也会被取消。对于上述情况，比较好的方法是让它成为 Flyout 作为 ContextFlyout 内容就用 ListView 取代 MenuFlyout. 更多内容请查看以下内容。

<SwipeControl>
    <SwipeControl.ContextFlyout>
        <Flyout Opening="OpenMusicMenuFlyout">
            <ListView>
                <ListView.ItemTemplate>
                    <DataTemplate>
                        <TextBlock Text="{Binding}"/>
                    </DataTemplate>
                </ListView.ItemTemplate>
            </ListView>
        </Flyout>
    </SwipeControl.ContextFlyout>
</SwipeControl>

代码背后

private async void OpenMusicMenuFlyout(object sender, object e)
{
    var items = new ObservableCollection<string>() { "One", "Two", "Three" };
    var flyout = sender as Flyout;
    var menu =  flyout.Content as ItemsControl;
    await TestMetod();
    menu.ItemsSource = items;
}