Question

为什么vb中sendmessage函数返回直总是0

保证过程全对．为什么vb中sendmessage函数返回直总是0？？？
Private Declare Function SendMessage Lib "user32" Alias "SendMessageA" (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, lParam As Any) As Long
Dim a, b, c, d As Long
Private Const WM_LBUTTONDOWN = &H201
Private Const WM_LBUTTONUP = &H202
请指点小第一二吧，我都弄了一天了就不行
Private Declare Function PostMessage Lib "user32" Alias "PostMessageA" (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, ByVal lParam As Long) As Long
Private Declare Function FindWindow Lib "user32" Alias "FindWindowA" (ByVal lpClassName As String, ByVal lpWindowName As String) As Long
Private Declare Function FindWindowEx Lib "user32" Alias "FindWindowExA" (ByVal hWnd1 As Long, ByVal hWnd2 As Long, ByVal lpsz1 As String, ByVal lpsz2 As String) As Long

Private Sub Command1_Click()
a = FindWindow("ThunderRT6FormDC", "Form1")
b = FindWindowEx(a, 0, "ThunderRT6CommandButton", "确定")
c = sendMessage(b, WM_LBUTTONDOWN, 0, 0)
d = sendMessage(b, WM_LBUTTONUP, 0, 0)

End Sub

Answer 1

Private Declare Function SendMessage Lib "user32" Alias "SendMessageA" (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, lParam As Any) As Long
Private Const WM_LBUTTONDOWN = &H201
Private Const WM_LBUTTONUP = &H202
Private Declare Function PostMessage Lib "user32" Alias "枝困慎PostMessageA" (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, ByVal lParam As Long) As Long
Private Declare Function FindWindow Lib "user32"猛敬尺氏 Alias "FindWindowA" (ByVal lpClassName As String, ByVal lpWindowName As String) As Long
Private Declare Function FindWindowEx Lib "user32" Alias "FindWindowExA" (ByVal hWnd1 As Long, ByVal hWnd2 As Long, ByVal lpsz1 As String, ByVal lpsz2 As String) As Long

Private Sub Command1_Click()
Dim a As Long, b As Long, c As Long, d As Long
a = FindWindow("ThunderRT6FormDC", "Test")
b = FindWindowEx(a, 0, "ThunderRT6CommandButton", "确定")
c = SendMessage(b, WM_LBUTTONDOWN, 0, ByVal 0)
d = SendMessage(b, WM_LBUTTONUP, 0, ByVal 0)
End Sub

Answer 2

你贴出你的程序？