Question

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，我已成功使用jni4net生成Java和.NET代码之间接口的代理。

此集成会在显示WPF UI时产生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()

通过使用以下模式使用ShowDialog加载WPF UI来克服这个问题：

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

然而，现在，我遇到类似于以下的异常，同时使用WPF UI，鼠标单击或按键可以触发以下内容（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，并从堆栈跟踪中找出问题的原因。

非常感谢任何帮助/建议。

Answer 1

我无法就解决方案提供太多建议，但我可以告诉你这个问题。您的所有WPF代码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有人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API ShowDialog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，以便将这些调用封送到相应的WPF（UI）线程。

所以它应该做类似的事情：

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

不幸的是，这可能意味着你将不得不在你的API上做很多工作，因为它可能不会解决像这样的线程问题。

在Java应用程序中使用WPF库时遇到问题？

1 个答案: