我如何从启动它的进程中激活一个正在运行的进程的窗口? [英] How do I activate the window of one running process from the process which started it?

问题描述

我已经开始了一个过程:

I have started a process:

dim myproc as new process
'' Some setup code for calling a third-party application
myproc.start

在程序的后面，我需要将myproc的主窗口设置为最顶层的活动窗口.

有什么建议?我无法修改第三方应用程序.

根据一个建议，我尝试了:

Later in the program, I need to make the main window of myproc the topmost, active window.

Any suggestions? I cannot modify the third-party application.

Per one suggestion, I tried:

'' Activate and bring to the top an existing window.
Dim winPtr As IntPtr

winPtr = myproc.MainWindowHandle
WinAPI.SendMessage(winPtr, WinAPI.WndMsg.WM_SETFOCUS, CType(1, IntPtr), CType(0, IntPtr))

但这不起作用.

but this did not work.

推荐答案

您需要获取其他应用程序窗口的句柄，并向该窗口发送WM_FOCUS 消息.

如果您自己启动了该应用程序，则可以通过操纵用于启动该应用程序的Process对象来获取窗口句柄.

You need to get the handle to the other applications window, and send that window a WM_FOCUS message.

If you launched the app yourself, you could get the window handle by way of manipulating the Process object you used to launch the application.

该过程变量必须定义如下:

The process variable must be defined as follows:

Private Shared WithEvents myProcess As Process = Nothing

然后，您可以捕获Exited事件.效果很好.

Then you can catch the Exited event. This works beautifully.

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