IMO的答案错误,它建议你启动一个线程来处理此问题。你需要做的是将窗口跳转回主调度线程。
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;
}
public Dispatcher UIDispatcher { get; private set; }
然后,当另一个线程(在这种情况下是线程池线程)触发了某个事件:
public void Consume(ImageFound message)
{
var model = _container.Resolve<ChoiceViewModel>();
model.ForImage(message);
UIDispatcher.BeginInvoke(new Action(() => _windows.ShowWindow(model)));
}
WinForms等效功能
不要将UIDispatcher设置为任何值,这样您就可以进行以下操作:
public void Consume(ImageFound message)
{
var model = _container.Resolve<ChoiceViewModel>();
model.ForImage(message);
this.Invoke( () => _windows.ShowWindow(model) );
}
优化WPF代码的DRY原则:
哇,这么多代码……
public interface ThreadedViewModel
: IConsumer
{
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);
}
}
在视图模型中:
public void Consume(ImageFound message)
{
var model = _container.Resolve<ChoiceViewModel>();
model.ForImage(message);
this.BeginInvoke(() => _windows.ShowWindow(model));
}
希望这有所帮助。