- This topic has 6 replies, 5 voices, and was last updated 18 years ago by Newbie.
Viewing 7 posts - 1 through 7 (of 7 total)
-
AuthorPosts
-
October 22, 2006 at 6:54 pm #190970ii Rocky iiMember
what is problem in this code
Private Declare Sub keybd_event Lib "user32" (ByVal bVk As Byte, ByVal bScan As Byte, ByVal dwFlags As Long, ByVal dwExtraInfo As Long)Pivate Const KEYEVENTF_KEYUP = &H2 Private Const VK_F9 = &H78 Private Sub Command1_Click() Call keybd_event(VK_F9, 0, 0, 0) End Sub Private Sub Command2_Click() Call keybd_event(VK_F9, 0, KEYEVENTF_KEYUP, 0) end Sub
October 22, 2006 at 8:11 pm #190976NewbieMember@ii Rocky ii wrote:
what is problem in this code
Private Declare Sub keybd_event Lib "user32" (ByVal bVk As Byte, ByVal bScan As Byte, ByVal dwFlags As Long, ByVal dwExtraInfo As Long)Pivate Const KEYEVENTF_KEYUP = &H2 Private Const VK_F9 = &H78
should be
Private Declare Sub keybd_event Lib "user32" (ByVal bVk As Byte, ByVal bScan As Byte, ByVal dwFlags As Long, ByVal dwExtraInfo As Long) Private Const KEYEVENTF_KEYUP = &H2 Private Const VK_F9 = &H78
October 22, 2006 at 8:30 pm #190975ii Rocky iiMemberthx mistake is pivate
October 23, 2006 at 9:13 am #190974BattleStar-GalacticaMemberdouble posting should be killed 😆
October 23, 2006 at 1:44 pm #190973AdminAdministratori second that^
@nanomachine007 wrote:
double posting should be killed 😆
November 20, 2006 at 5:42 pm #190972Sabotage2Memberwhy double posting willbee killed 😯
November 20, 2006 at 6:17 pm #190971AdminAdministratorits a leeched code from planet-source-code.com
@ii Rocky ii wrote:
what is problem in this code
Private Declare Sub keybd_event Lib "user32" (ByVal bVk As Byte, ByVal bScan As Byte, ByVal dwFlags As Long, ByVal dwExtraInfo As Long)Pivate Const KEYEVENTF_KEYUP = &H2 Private Const VK_F9 = &H78 Private Sub Command1_Click() Call keybd_event(VK_F9, 0, 0, 0) End Sub Private Sub Command2_Click() Call keybd_event(VK_F9, 0, KEYEVENTF_KEYUP, 0) end Sub
-
AuthorPosts
Related
Viewing 7 posts - 1 through 7 (of 7 total)
- You must be logged in to reply to this topic.