c# - Task.Run() 的回调方法

Question

目前我的代码变得非常重复。我必须在整个软件过程中反复提高忙碌指示器。

这三个 Action 是

1. Raise Busy Indicator
2. Do the actions
3. Turn Off Busy Indicator

例子

public async void OpenAttachment()
{
    Events.PublishOnUIThread(new BusyEvent { IsBusy = true });
    await Task.Run(() =>
    {
        try
        {
            if (SelectedAttachment == null)
            {
                return;
            }

            var tempFile = string.Format(
                "{0}\\{1}.{2}", Path.GetTempPath(), SelectedAttachment.FileName, SelectedAttachment.FileExtension);

            System.IO.File.WriteAllBytes(tempFile, UnitOfWork.FileRepository.GetFileBytes(SelectedAttachment.Id));

            Process.Start(tempFile);
        }
        catch (Exception ex)
        {
            Notification.Error("Person - Opening attachment", "File couldn't open, please close last file instance.");
        }
    });
    Events.PublishOnUIThread(new BusyEvent { IsBusy = false });
}

我希望运行一种方法，这样它就可以执行繁忙指示器，而不必每次都重复它。

有点像

public async void OpenAttachment()
{
    Execute(() => await Task.Run(() => {....TaskWork});
}

想知道是否有人可以给我提示如何减少这些重复代码。

Answer 1

你的意思是这样的？

public async Task RunBusyTask(Action task)
{
    Events.PublishOnUIThread(new BusyEvent { IsBusy = true });
    await Task.Run(task);
    Events.PublishOnUIThread(new BusyEvent { IsBusy = false });
}

---

RunBusyTask(() => {...});