C# 在 BackgroundWorker 中调用 ShowDialog
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Calling ShowDialog in BackgroundWorker
提问by Ahmed
I have a WinForms application in which my background worker is doing a sync task, adding new files, removing old ones etc.
我有一个 WinForms 应用程序，我的后台工作人员正在其中执行同步任务、添加新文件、删除旧文件等。
In my background worker code I want to show a custom form to user telling him what will be deleted and what will be added if he continues, with YES/NO buttons to get his feedback.
在我的后台工作代码中，我想向用户显示一个自定义表单，告诉他将删除哪些内容以及如果他继续操作将添加哪些内容，并使用“是”/“否”按钮来获取他的反馈。
I was wondering if it is ok to do something like this in background worker's doWork method? If not, how should I do it?
我想知道在后台工作人员的 doWork 方法中做这样的事情是否可以？如果没有，我该怎么做？
Please advise..
请指教..
private void backgroundWorker1_DoWork(object sender, DoWorkEventArgs e)
{
MyForm f = new MyForm();
f.FilesToAddDelete(..);
DialogResult result = f.ShowDialog();
if(No...)
return;
else
//keep working...
}
采纳答案by Jon
If you try this you will see for yourself that it will not work because the BackgroundWorkerthread is not STA(it comes from the managed thread pool).
如果您尝试这样做，您会亲眼看到它不会工作，因为该BackgroundWorker线程不是STA（它来自托管线程池）。
The essence of the matter is that you cannot show user interface from a worker thread1, so you must work around it. You should pass a reference to a UI element of your application (the main form would be a good choice) and then use Invoketo marshal a request for user interaction to your UI thread. A barebones example:
问题的本质是您无法从工作线程 1 显示用户界面，因此您必须解决它。您应该传递对应用程序 UI 元素的引用（主窗体将是一个不错的选择），然后用于Invoke将用户交互请求编组到您的 UI 线程。一个准系统示例：
class MainForm
{
// all other members here
public bool AskForConfirmation()
{
var confirmationForm = new ConfirmationForm();
return confirmationForm.ShowDialog() == DialogResult.Yes;
}
}
And the background worker would do this:
后台工作人员会这样做：
// I assume that mainForm has been passed somehow to BackgroundWorker
var result = (bool)mainForm.Invoke(mainForm.AskForConfirmation);
if (result) { ... }
1 Technically, you cannot show user interface from a thread that is not STA. If you create a worker thread yourself you can choose to make it STA anyway, but if it comes from the thread pool there is no such possibility.
1 从技术上讲，您不能从非 STA 的线程显示用户界面。如果你自己创建一个工作线程，你可以选择让它成为 STA，但如果它来自线程池，则没有这种可能性。
回答by Henrik
IMO answers stating that you should launch a thread to handle this are misguided. What you need is to jump the window back to the main dispatcher thread.
IMO 回答说你应该启动一个线程来处理这个问题是被误导的。您需要的是将窗口跳转回主调度程序线程。
In WPF
在 WPF 中
public ShellViewModel(
[NotNull] IWindowManager windows,
[NotNull] IWindsorContainer container)
{
if (windows == null) throw new ArgumentNullException("windows");
if (container == null) throw new ArgumentNullException("container");
_windows = windows;
_container = container;
UIDispatcher = Dispatcher.CurrentDispatcher; // not for WinForms
}
public Dispatcher UIDispatcher { get; private set; }
and then, when some event occurs on another thread (thread pool thread in this case):
然后，当另一个线程（在本例中为线程池线程）上发生某个事件时：
public void Consume(ImageFound message)
{
var model = _container.Resolve<ChoiceViewModel>();
model.ForImage(message);
UIDispatcher.BeginInvoke(new Action(() => _windows.ShowWindow(model)));
}
WinForms equivalent
WinForms 等效项
Don't set UIDispatcher to anything, then you can do have:
不要将 UIDispatcher 设置为任何内容，然后您可以：
public void Consume(ImageFound message)
{
var model = _container.Resolve<ChoiceViewModel>();
model.ForImage(message);
this.Invoke( () => _windows.ShowWindow(model) );
}
DRYing it up for WPF:
为 WPF 干燥：
Man, so much code...
伙计，这么多代码......
public interface ThreadedViewModel
: IConsumer
{
/// <summary>
/// Gets the UI-thread dispatcher
/// </summary>
Dispatcher UIDispatcher { get; }
}
public static class ThreadedViewModelEx
{
public static void BeginInvoke([NotNull] this ThreadedViewModel viewModel, [NotNull] Action action)
{
if (viewModel == null) throw new ArgumentNullException("viewModel");
if (action == null) throw new ArgumentNullException("action");
if (viewModel.UIDispatcher.CheckAccess()) action();
else viewModel.UIDispatcher.BeginInvoke(action);
}
}
and in the view model:
并在视图模型中：
public void Consume(ImageFound message)
{
var model = _container.Resolve<ChoiceViewModel>();
model.ForImage(message);
this.BeginInvoke(() => _windows.ShowWindow(model));
}
Hope it helps.
希望能帮助到你。
回答by Tomtom
You should bring up the dialog before you run the backgroundworker. And in the progresschanged-event, you can update the dialog.
您应该在运行后台工作器之前打开对话框。在progresschanged-event 中，您可以更新对话框。
回答by C-Pound Guru
I usually create a method to execute a delegate on the UI thread:
我通常会创建一个方法来在 UI 线程上执行委托：
private void DoOnUIThread(MethodInvoker d) {
if (this.InvokeRequired) { this.Invoke(d); } else { d(); }
}
With this, you can change your code to:
有了这个，您可以将代码更改为：
private void backgroundWorker1_DoWork(object sender, DoWorkEventArgs e)
{
DialogResult result = DialogResult.No;
DoOnUIThread(delegate() {
MyForm f = new MyForm();
f.FilesToAddDelete(..);
result = f.ShowDialog();
});
if(No...)
return;
else
//keep working...
}
