Question

VB这段代码如何修改，可以双击托盘图标打开程序。

窗体上创建一个菜单
标题为“文件” 名称为sys的菜单 设置不可见
有两个子菜单 “退出” NMexit
“显示” NMshow
Option Explicit
Const MAX_TOOLTIP As Integer = 60
Const NIF_ICON = &H2
Const NIF_MESSAGE = &H1
Const NIF_TIO = &H4
Const NIM_ADD = &H0
Const NIM_DELETE = &H2
Const WM_MOUSEMOVE = &H200
Const WM_LBUTTONUP = &H202
Const WM_RBUTTONUP = &H205
Private Type NOTIFYICONDATA
cbSize As Long
bwnd As Long
uID As Long
uFlags As Long
uCallBackMessage As Long
hIcon As Long
szTip As String * MAX_TOOLTIP
End Type
'声明 Shell_NotifyIcon 函数
Private Declare Function Shell_NotifyIcon Lib "shell32.dll" Alias "Shell_NotifyIconA" _
(ByVal dwMessage As Long, IpData As NOTIFYICONDATA) As Long
Private nfIconData As NOTIFYICONDATA

Private Sub Form_Load()
nfIconData.bwnd = Me.hWnd
nfIconData.uID = Me.Icon
nfIconData.uFlags = NIF_ICON Or NIF_MESSAGE Or NIF_TIO
nfIconData.uCallBackMessage = WM_MOUSEMOVE
nfIconData.hIcon = Me.Icon.Handle
nfIconData.szTip = "用vb制作的 QQ" & vbNullChar
nfIconData.cbSize = Len(nfIconData)
Call Shell_NotifyIcon(NIM_ADD, nfIconData)
End Sub
'窗体鼠标移动时间
Private Sub Form_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)
Dim IMsg As Single
IMsg = X / Screen.TwipsPerPixelX
If IMsg = WM_RBUTTONUP Or IMsg = WM_RBUTTONUP Then Me.PopupMenu sys
End Sub
'退出菜单
Private Sub NMexit_Click()
End
End Sub
'现实菜单
Private Sub NMshow_Click()
Me.Visible = True
Me.WindowState = 0
End Sub
如何双击托盘图标打开程序，求高手帮助。

Answer 1

Private Sub Form_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)
Dim IMsg As Single
IMsg = X / Screen.TwipsPerPixelX
If IMsg = WM_RBUTTONUP Or IMsg = WM_RBUTTONUP Then Me.PopupMenu sys
'加下面一句！
If IMsg = &H203 Then Me.Show
End Sub
请参阅我的博客：http://hi.baidu.com/zgmg/blog/item/1b965109731ffcb12eddd4a5.html