Question

vb中怎样才能双击托盘图标后直接显示程序？

vb程序最小化到托盘后鼠标双击托盘图标程序最小化到任务栏里，请问怎样才能双击后直接显示程序？
我是按你的代码做的
http://zhidao.baidu.com/question/434582369.html

Answer 1

Case WM_LBUTTONDBLCLK
Me.WindowState = 0 '加一句代码就可以了
Me.Show

追问

在哪儿加？

追答

Private Sub Form_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)
Static Message As Long

'x is the current mouse location along the x-axis
Message = X / Screen.TwipsPerPixelX

If RR = False Then
RR = True
Select Case Message
' Left double click (This should bring up a dialog box)
Case WM_LBUTTONDBLCLK
Me.Show '在这里加啊
' Right button up (This should bring up a menu)
Case WM_RBUTTONUP
' Me.PopupMenu mnuPopUp
End Select
RR = False
End If

End Sub

追问

ok了，非常感谢您。顺便帮我解决一下退出程序后图标还在托盘里 的问题，鼠标移动到图标才会消失。

Answer 2

在窗口的双击事件中添加代码
me.show