在vb6中动态生成两个或更多级别的子菜单

时间:2009-05-23 17:21:33

标签: vb6 dynamic creation submenu

友 告诉我如何在运行时在VB6中生成超过1级子菜单?简要解释一下?有没有具体的控制?但我不想使用外部控件!

2 个答案:

答案 0 :(得分:4)

您可以使用API​​函数

创建多个子菜单级别
Private Declare Function CreatePopupMenu Lib "user32" () As Long

Private Declare Function AppendMenu Lib "user32" Alias "AppendMenuA" (ByVal hmenu As Long, ByVal wFlags As Long, ByVal wIDNewItem As Long, ByVal lpNewItem As Any) As Long

Private Declare Function GetCursorPos Lib "user32" (lpPoint As POINTAPI) As Long

Private Declare Function TrackPopupMenu Lib "user32" (ByVal hmenu As Long, ByVal wFlags As Long, ByVal X As Long, ByVal Y As Long, ByVal nReserved As Long, ByVal hwnd As Long, lprc As Any) As Long

Private Declare Function DestroyMenu Lib "user32" (ByVal hmenu As Long) As Long

Private Type POINTAPI
    X As Long
    Y As Long
End Type

Dim hmenu As Long, hSubMenu As Long
Private Const MF_STRING = &H0&
Private Const MF_SEPARATOR = &H800&


  hSubMenu = CreatePopupMenu
  AppendMenu hSubMenu, 0, 121, "Sub Menu1"
  AppendMenu hSubMenu, 0, 122, "Sub Menu2"

  hmenu = CreatePopupMenu
  AppendMenu hmenu, 0, 107, "Menu1"

  AppendMenu hmenu, 0, 106, "Menu2"

  AppendMenu hmenu, MF_POPUP, hSubMenu, "Menu3"
  AppendMenu hmenu, MF_POPUP, hSubMenu, "Menu4"

  AppendMenu hmenu, 0, 101, "Menu5"

显示

  If Button = vbRightButton Then
    Dim P As POINTAPI
    GetCursorPos P
    TrackPopupMenu hmenu, 0, P.X, P.Y, 0, hwnd, 0

在调用TrackPopupMenu之前,不会显示菜单。其返回值可以指示选择了哪个(如果有)菜单项。例如,如果选择“Menu1”,它可以返回'107'。

答案 1 :(得分:2)

您可以使用标准VB菜单执行此操作,但由于您必须使用控件数组,因此您必须在设计时创建一个带Index = 0的第一个原型菜单(例如mnuFoo(0))(通常是看不见的)。您现在可以动态加载新项目。

Call Me.Load(mnuFoo(1)) ' New array member (index 1) '
With mnuFoo(1)
    .Visible = True ' Make it visible
    ' --- Do some settings
End With