How do I activate the window of one running process from the process which started it?

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I have started a process:

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

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))

but this did not work.

Posted 15-Mar-10 10:26am

Jalapeno Bob

Updated 16-Mar-10 10:18am

v9

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2 solutions

Solution 2

The process variable must be defined as follows:

Private Shared WithEvents myProcess As Process = Nothing

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

Posted 26-Apr-10 7:26am

Jalapeno Bob

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#realJSOP · Accepted Answer · 2010-03-15T11:59:00

Solution 1

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.

Posted 15-Mar-10 11:59am

#realJSOP

Updated 15-Mar-10 12:32pm

v3