激活 WPF 窗口而不会失去对以前的应用程序/窗口的关注
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Activate WPF Window without losing focus on previous application/window
提问by deathrace
I have a WPF window in application which I Activate on some specific scenarios by calling MainView.Activate();and MainView.BringIntoView();method. it also sets focus on this 'MainView' window.
我在应用程序中有一个 WPF 窗口，我通过调用MainView.Activate();和MainView.BringIntoView();方法在某些特定场景中激活它。它还将焦点设置在这个“MainView”窗口上。
But my requirement is this window should not get Focus. i.e. my cursor should still remain on previous application(notepad,word etc..)
但我的要求是这个窗口不应该获得焦点。即我的光标应该仍然保留在以前的应用程序（记事本、word 等）上
I tried using MainView.ShowActivated="False"but it didn't work.
Do I need to use HwndSource as mentioned hereor what?
我尝试使用MainView.ShowActivated="False"但它没有用。我需要使用这里提到的 HwndSource还是什么？
Code I have used after Kent's help (Its working only if Window is minimized):
我在 Kent 的帮助下使用的代码（仅当窗口最小化时才有效）：
IntPtr HWND_TOPMOST = new IntPtr(-1);
const short SWP_NOMOVE = 0X2;
const short SWP_NOSIZE = 1;
const short SWP_NOZORDER = 0X4;
const int SWP_SHOWWINDOW = 0x0040;
Process[] processes = Process.GetProcesses(".");
foreach (var process in processes)
{
IntPtr handle = process.MainWindowHandle;
if (handle != IntPtr.Zero)
{
SetWindowPos(handle, HWND_TOPMOST, 0, 0, 0, 0, SWP_NOZORDER | SWP_NOSIZE | SWP_SHOWWINDOW);
}
}
回答by Kent Boogaart
In my recent blog post, I use SetWindowPosto bring a window to the front of the z-order without giving it focus. I don't believe WPF has an in-built means of achieving the same without p/invoke.
在我最近的博客文章中，我习惯于SetWindowPos将一个窗口放在 z-order 的前面而不给它焦点。我认为 WPF 没有内置的方法可以在没有 p/invoke 的情况下实现相同的目标。
回答by Eli Dagan
Let call the window that you want to show without focus: YourWindow. Add the following code:
让我们调用您想要在没有焦点的情况下显示的窗口：YourWindow。添加以下代码：
YourWindow.ShowActivted = false;
YourWindow.Owner = TheOtherWindowThatWillStilHaveFocus;
YourWindow.ShowActivted = false;
YourWindow.Owner = TheOtherWindowThatWillStilHaveFocus;
First line you can also be set from Xaml.
也可以从 Xaml 设置第一行。
回答by Mikhail
I Have Win32 helper class for such situation here is listing for your case
我有针对这种情况的 Win32 助手类，这里列出了您的案例
using System;
using System.Runtime.InteropServices;
using System.Windows;
using System.Windows.Interop;
namespace your.namespace {
public static class Win32 {
public static void ShowWindowNoActive( Window window) {
var hwnd = (HwndSource.FromVisual(window) as HwndSource).Handle;
ShowWindow(hwnd, ShowWindowCommands.SW_SHOWNOACTIVATE);
}
[DllImport("user32.dll")]
private static extern bool ShowWindow(IntPtr hWnd, ShowWindowCommands nCmdShow);
private enum ShowWindowCommands : int {
SW_SHOWNOACTIVATE = 4
}
}
}
回答by ngm
I needed to solve a similar problem and got it to work by setting Window.ShowActivate=false and then using Window.Show rather than Window.Activate. Perhaps this will work for you. I have not yet needed to use BringIntoView because my window is newly created and appears on top of other windows by default.
我需要解决一个类似的问题，并通过设置 Window.ShowActivate=false 然后使用 Window.Show 而不是 Window.Activate 来让它工作。也许这对你有用。我还不需要使用BringIntoView，因为我的窗口是新创建的，默认情况下显示在其他窗口的顶部。
