UI线程上的异步WPF多状态更新

Question

我想从一个长期运行的方法状态更新。通常我会使用dispatcher回发到UI线程，但我对使用async Await很好奇。

为了简单起见：

<Button Name="ButtonWithCodeBehind" Height="25" Click="ButtonWithCodeBehindOnClick"/>

private async void ButtonWithCodeBehindOnClick(object sender, RoutedEventArgs e)
{
    await Task.Factory.StartNew(() =>
    {
        ButtonWithCodeBehind.Content = "First";
        Thread.Sleep(1000);
        ButtonWithCodeBehind.Content = "Second";
        Thread.Sleep(1000);
        ButtonWithCodeBehind.Content = "Third";
    });
}

Deployment.Current.Dispatcher.BeginInvoke(()=>ButtonWithCodeBehind.Content = "Second");

private async void ButtonWithCodeBehindOnClick(object sender, RoutedEventArgs e)
{
    var scheduler = TaskScheduler.FromCurrentSynchronizationContext();
    await Task.Factory.StartNew(() =>
    {
        Task.Factory.StartNew(() => Thread.Sleep(1000))
            .ContinueWith(t => ButtonWithCodeBehind.Content = "First", scheduler)
            .ContinueWith(t => Thread.Sleep(1000))
            .ContinueWith(t => ButtonWithCodeBehind.Content = "Second", scheduler)
            .ContinueWith(t => Thread.Sleep(1000))
            .ContinueWith(t => ButtonWithCodeBehind.Content = "Third", scheduler);
    });
}

但这是Fugly。此外，如果您去掉async和await关键字并将其替换为task.waitall，它仍将按预期执行。

注意：如果您想知道为什么我使用thread.sleep而不是task.delay，我实际上也在Silverlight中测试了这一点，异步等待支持不包括。delay（或者至少不包括我期望的位置）。

Answer 1

如果您可以将长时间运行的任务拆分为两个不同的长时间运行的操作（就像上面示例中的两个Thread.Sleeps)，您可以单独等待每个长时间运行的任务。因此UI更新将在UI线程上执行。

private async void ButtonWithCodeBehindOnClick(object sender, RoutedEventArgs e)
{
  ButtonWithCodeBehind.Content = "First";
  await Task.Run(() => Thread.Sleep(1000));
  ButtonWithCodeBehind.Content = "Second";
  await Task.Run(() => Thread.Sleep(1000));
  ButtonWithCodeBehind.Content = "Third";
}