How can I tell if a menu is open in VB6?

Question

I've got a timer set up to detect if the mouse is over a certain area on my form, which you can imagine to be a rectangle starting at 50,50 (pixels) and ending at 1000,500. If the mouse is inside that rectangle, a second window pops up that acts somewhat like a tooltip, following the mouse around. The problem is that the menus at the top drape over this rectangle, and if you try to use a menu, the second window pops up (the timer sets its visible property to true) as soon as you move down the menu, which ends up closing the menu (I guess due to a loss of focus or something.)

If I can detect when one of the menus is open, I can disable the showing of the tooltip window with an if statement, but I don't know how to do that.

score 0 · Accepted Answer · answered Jul 04 '12 at 06:53

I think I've figured out how to do this by searching WIN32API.txt for "menu" and a little bit of googling, but I'm not really sure. Perhaps this solution only works on my machine.

Putting this code...

Dim hMenu As Long
hMenu = GetMenu(Form1.hwnd)
MsgBox GetMenuState(hMenu, 0, MF_BYPOSITION)

on a timer with an interval of 5000 allows you to the view the menu's state. In a closed state, the number appears to be random (1552, 1296, etc.,) but when the menu is opened, it is offset by 128 from this base value. A menu whose state is 1552 when closed is 1680 when open.

I'm not sure why it's offset by 128 or if this works on all machines (just to be safe, I will program it to check for inequality, not offset by 128), but it appears to be working for me.

If there is a problem with this solution or if there is a better way, please respond with another answer, and I'll be glad to give you credit for the answer instead.

The state is a bit field. To see this, convert the number to binary in a calculator, then research bit masks. — Deanna, Jul 04 '12 at 10:06
@Deanna Queen of all things VB6, do you know the answer to this question? http://stackoverflow.com/questions/11515724/attempting-to-learn-polymorphism-etc-in-vb6-but-my-code-doesnt-do-what-i-wan/ — TimFoolery, Jul 17 '12 at 08:29
@Deanna Disregard; we figured it out. Severe case of user error. :S — TimFoolery, Jul 17 '12 at 17:34

user14938123 · Answer 2 · 2021-01-06T10:10:43.003

Written by SBEIH Iyad - Syria - Damascus. 4/1/2021.

Using VB6.0, we can. We check all menus windows if one is opened, it means: menu is opened. we use API "GetMenu" for hWnd of main-menu, then with the API "GetMenuState" we check if one of menus is opened.

Vb6.0 CODE:

Private Function GetBit_I(ByVal X As Long, ByVal i As Integer) As Integer
' Get the bit number i of X.
  GetBit_I = (X And (2 ^ i)) / (2 ^ i)
End Function

Private Function MainMenuIsOpened(FRM As Form) As Boolean
Const MF_BYPOSITION = 1024
Dim H As Long, i As Integer, L As Long, MCount As Long
  MainMenuIsOpened = False
  On Error GoTo MainMenuIsOpenedError
  H = GetMenu(FRM.HWnd)
  MCount = GetMenuItemCount(H)
  ' MCount is the number of main-menus.
  Do While (i < MCount)
     L = GetMenuState(H, i, MF_BYPOSITION)
     If ((L > -1) And (GetBit_I(L, 7) = 1)) Then
        MainMenuIsOpened = True
        Exit Do
     End If
     i = i + 1
  Loop
  Exit Function
MainMenuIsOpenedError:
  MainMenuIsOpened = False
End Function

Good luck.

How can I tell if a menu is open in VB6?

2 Answers2