C# 使用窗口句柄在最上面创建一个窗口
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Make a window topmost using a window handle
提问by James Cadd
After launching an application using the Process class I'd like to make that window topmost. Currently, my app is the topmost window so when i launch the other app it doesn't display. One thing that came to mind is that I could set topmost = false for my application before launching the process, the problem with this is I want to give the process ample time to load up before displaying it to the user, so I'd like more control over when I switch the other application to the topmost.
使用 Process 类启动应用程序后，我想让该窗口位于最顶部。目前，我的应用程序是最顶层的窗口，所以当我启动另一个应用程序时它不会显示。我想到的一件事是我可以在启动进程之前为我的应用程序设置 topmost = false，这个问题是我想在向用户显示它之前给进程足够的时间来加载它，所以我想当我将另一个应用程序切换到最顶层时，可以更好地控制。
采纳答案by Reed Copsey
You need to use P/Invoke with SetWindowPosto accopmlish this:
你需要使用P/Invoke 和 SetWindowPos来完成这个：
[DllImport("user32.dll")]
static extern bool SetWindowPos(IntPtr hWnd, IntPtr hWndInsertAfter, int X, int Y, int cx, int cy, uint uFlags);
static readonly IntPtr HWND_TOPMOST = new IntPtr(-1);
const UInt32 SWP_NOSIZE = 0x0001;
const UInt32 SWP_NOMOVE = 0x0002;
const UInt32 SWP_SHOWWINDOW = 0x0040;
// Call this way:
SetWindowPos(theWindowHandle, HWND_TOPMOST, 0, 0, 0, 0, SWP_NOMOVE | SWP_NOSIZE | SWP_SHOWWINDOW);
