发现WPF中一个关于TouchDown事件和MouseDown事件的激活窗口的问题。
0
悬赏园豆：5
[待解决问题]
TrayWindow和MenuWindow两个窗口，程序启动时都显示。
在鼠标点击一下TrayWindow或者在触摸屏上触碰一下TrayWindow，如果MenuWindow是显示的就隐藏，如果是隐藏的就显示。
此外，MenuWindow在Deactivated时也会隐藏自己。
代码如下
App.xaml
<Application x:Class="TouchTest.App" xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation" xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml" Startup="Application_Startup"> </Application>
App.xaml.cs
using System; using System.Windows; namespace TouchTest { public partial class App : Application { private void Application_Startup(object sender, StartupEventArgs e) { MenuWindow menuWin = new MenuWindow(); TrayWindow trayWin = new TrayWindow(menuWin); trayWin.Show(); menuWin.Show(); } } }
MenuWindow.xaml
<Window x:Class="TouchTest.MenuWindow" xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation" xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml" Title="Menu" Height="300" Width="300" Deactivated="Window_Deactivated"> </Window>
MenuWindow.xaml.cs
using System; using System.Windows; using System.Windows.Controls; using System.Diagnostics; namespace TouchTest { public partial class MenuWindow : Window { public MenuWindow() { InitializeComponent(); } private void Window_Deactivated(object sender, EventArgs e) { Debug.WriteLine("Hide By Window_Deactivated"); Hide(); } } }
TrayWindow.xaml
<Window x:Class="TouchTest.TrayWindow" xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation" xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml" Title="Tray" Height="202" Width="279" TouchDown="Window_TouchDown" MouseDown="Window_MouseDown" Closed="Window_Closed"> </Window>
TrayWindow.xaml.cs
using System; using System.Windows; using System.Windows.Controls; using System.Windows.Shapes; using System.Diagnostics; using System.Windows.Input; using System.Windows.Interop; namespace TouchTest { public partial class TrayWindow : Window { private MenuWindow m_MenuWindow; public TrayWindow(MenuWindow menuWindow) { InitializeComponent(); m_MenuWindow = menuWindow; } private void Window_TouchDown(object sender, TouchEventArgs e) { if (m_MenuWindow.Visibility == Visibility.Visible) { Debug.WriteLine("Hide By Window_TouchDown"); m_MenuWindow.Hide(); } else { Debug.WriteLine("Show By Window_TouchDown"); m_MenuWindow.Show(); } e.Handled = true; } private void Window_MouseDown(object sender, MouseButtonEventArgs e) { if (m_MenuWindow.Visibility == Visibility.Visible) { Debug.WriteLine("Hide By Window_MouseDown"); m_MenuWindow.Hide(); } else { Debug.WriteLine("Show By Window_MouseDown"); m_MenuWindow.Show(); } } private void Window_Closed(object sender, EventArgs e) { Application.Current.Shutdown(); } } }
问是通过鼠标点击可以正常的显示隐藏MenuWindow.
但是MenuWindow显示后，在触摸屏上点击TrayWindow，MenuWindow会闪一下，不会隐藏。
经过调试发现，MenuWindow显示后，在TrayWindow中点击鼠标时，MenuWindow先被Deactivated, 此时MenuWindow就被隐藏，而且TrayWindow收不到MouseDown事件。
而在MenuWindow显示后，通过触摸屏点击TrayWindow，MenuWindow也是先被Deactivated，此时MenuWindow也被隐藏了，但是紧接着TrayWindow收到了TouchDown事件，由于此时MenuWindow已经不可见，TouchDown处理函数就把MenuWindow显示出来。这也是MenuWindow闪一下的原因。
经过查看MSDN中WM_MOUSEACTIVATE(https://msdn.microsoft.com/EN-US/library/windows/desktop/ms645612%28v=vs.85%29.aspx)的介绍，在TrayWindow中加入如下的代码，拦截WM_MOUSEACTIVATE，但是没有效果。
protected override void OnSourceInitialized(System.EventArgs e) { HwndSource source = PresentationSource.FromVisual(this) as HwndSource; if (source != null) source.AddHook(WndProc); } private IntPtr WndProc(IntPtr hwnd, int msg, IntPtr wParam, IntPtr lParam, ref bool handled) { if (msg != 0x0021) { return IntPtr.Zero; } else { handled = true; return new IntPtr(2); } }
有什么办法让MouseDown消息和TouchDown消息的效果一致。
所有回答(1)
0
应该用TouchUp吧？代表整个触摸动作已经完成。另外可以自己设置一个Flag去处理点击MenuWindow时另外一个窗口变成Deactivated的情况呀。