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我一直在尝试将现有的 C# WPF API 与 Java 应用程序连接起来。

到目前为止，我已经成功地使用jni4net生成代理来连接 Java 和 .NET 代码。

此集成在显示 WPF UI 时产生了 STA 线程问题：

System.InvalidOperationException: The calling thread must be STA, because many UI components require this.
  at System.Windows.Input.InputManager..ctor()
  at System.Windows.Input.InputManager.GetCurrentInputManagerImpl()
  at System.Windows.Input.KeyboardNavigation..ctor()
  at System.Windows.FrameworkElement.FrameworkServices..ctor()
  at System.Windows.FrameworkElement.EnsureFrameworkServices()
  at System.Windows.FrameworkElement..ctor()
  at System.Windows.Controls.Control..ctor()
  at System.Windows.Window..ctor()

这是通过使用以下模式来克服使用 ShowDialog 加载 WPF UI 的：

Thread thread = new Thread(new ParameterizedThreadStart(ParameterizedMethodName));
thread.SetApartmentState(ApartmentState.STA);
thread.Start(parameter);
thread.Join();

但是，现在，我在使用 WPF UI 时遇到类似于以下的异常，鼠标单击或按键可以触发以下情况（此示例来自鼠标单击）：

System.InvalidOperationException: The calling thread cannot access this object because a different thread owns it.
  at System.Windows.Threading.Dispatcher.VerifyAccess()
  at System.Windows.DependencyObject.GetValue(DependencyProperty dp)
  at System.Windows.Input.InputBinding.get_Command()
  at System.Windows.Input.InputBindingCollection.FindMatch(Object targetElement, InputEventArgs inputEventArgs)
  at System.Windows.Input.CommandManager.TranslateInput(IInputElement targetElement, InputEventArgs inputEventArgs)
  at System.Windows.UIElement.OnMouseDownThunk(Object sender, MouseButtonEventArgs e)
  at System.Windows.Input.MouseButtonEventArgs.InvokeEventHandler(Delegate genericHandler, Object genericTarget)
  at System.Windows.RoutedEventArgs.InvokeHandler(Delegate handler, Object target)
  at System.Windows.RoutedEventHandlerInfo.InvokeHandler(Object target, RoutedEventArgs routedEventArgs)
  at System.Windows.EventRoute.InvokeHandlersImpl(Object source, RoutedEventArgs args, Boolean reRaised)
  at System.Windows.UIElement.RaiseEventImpl(DependencyObject sender, RoutedEventArgs args)
  at System.Windows.UIElement.RaiseTrustedEvent(RoutedEventArgs args)
  at System.Windows.UIElement.RaiseEvent(RoutedEventArgs args, Boolean trusted)
  at System.Windows.Input.InputManager.ProcessStagingArea()
  at System.Windows.Input.InputManager.ProcessInput(InputEventArgs input)
  at System.Windows.Input.InputProviderSite.ReportInput(InputReport inputReport)
  at System.Windows.Interop.HwndMouseInputProvider.ReportInput(IntPtr hwnd, InputMode mode, Int32 timestamp, RawMouseActions actions, Int32 x, Int32 y, Int32 wheel)
  at System.Windows.Interop.HwndMouseInputProvider.FilterMessage(IntPtr hwnd, WindowMessage msg, IntPtr wParam, IntPtr lParam, Boolean& handled)
  at System.Windows.Interop.HwndSource.InputFilterMessage(IntPtr hwnd, Int32 msg, IntPtr wParam, IntPtr lParam, Boolean& handled)
  at MS.Win32.HwndWrapper.WndProc(IntPtr hwnd, Int32 msg, IntPtr wParam, IntPtr lParam, Boolean& handled)
  at MS.Win32.HwndSubclass.DispatcherCallbackOperation(Object o)
  at System.Windows.Threading.ExceptionWrapper.InternalRealCall(Delegate callback, Object args, Int32 numArgs)
  at MS.Internal.Threading.ExceptionFilterHelper.TryCatchWhen(Object source, Delegate method, Object args, Int32 numArgs, Delegate catchHandler)
  at System.Windows.Threading.Dispatcher.InvokeImpl(DispatcherPriority priority, TimeSpan timeout, Delegate method, Object args, Int32 numArgs)
  at MS.Win32.HwndSubclass.SubclassWndProc(IntPtr hwnd, Int32 msg, IntPtr wParam, IntPtr lParam)
  at MS.Win32.UnsafeNativeMethods.DispatchMessage(MSG& msg)
  at System.Windows.Threading.Dispatcher.PushFrameImpl(DispatcherFrame frame)
  at System.Windows.Threading.Dispatcher.PushFrame(DispatcherFrame frame)
  at System.Windows.Window.ShowHelper(Object booleanBox)
  at System.Windows.Window.Show()
  at System.Windows.Window.ShowDialog()

我目前不确定如何解决此问题并从堆栈跟踪中确定问题的原因。

非常感谢任何帮助/建议。

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1 回答 1

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我不能就解决方案提供太多建议，但我可以告诉你问题所在。您的所有 WPF 代码都必须在同一个线程上运行（可能是您正在创建的那个）。您得到的错误是因为某些东西试图从不同的线程访问 WPF 控件。

因此，假设您有一个显示对话框的 API 调用，并且您的 java 代码调用MyApi.ShowDialog. 您的 APIShowDialog方法不能简单地调用MyDialog.ShowDialog()，因为对 UI 组件的调用将来自 java 线程。相反，您的 API 需要足够智能以将这些调用编组到适当的 WPF (UI) 线程。

所以它应该做这样的事情：

if(!CheckAccess())
    MyDialog.Dispatcher.BeginInvoke(DeleageToShowDialog);

不幸的是，这可能意味着您将不得不在您的 API 上做很多工作，因为它可能无法解决此类线程问题。

于 2012-08-24T15:04:35.440 回答