Task.Run（）的回调方法

Question

at the moment my code is getting very repetitive. I have to raise my busy indicator repeatedly throughout the software.

The three actions are

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

Example

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 });
}

I'm looking to run a method such that it'll execute the busy indicator without having to repeat it everytime.

Something like

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

Wondering if someone can give me tips on how to reduce this repeated code.

Answer 1

You mean something like this?

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

---

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