调用系统API移动控件的有关问题

调用系统API移动控件的问题
我也在研究移动控件的问题， Declare Function SendMessage Lib "user32 " Alias "SendMessageA " (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, lParam As Any) As Long
调用他时候，系统说无法找到DLL的入口呢，是不是XP 的User32和98的Use32不同呢


Declare Function SendMessage Lib "user32 " Alias "SendMessageA " (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, lParam As Any) As Long

Public Const WM_SYSCOMMAND = &H112
Public Const SC_MOVE = &HF012

Private Sub Command2_MouseDown(Button As Integer, Shift As Integer, X As Single, Y As Single)
If Button = vbLeftButton Then
  '为当前的应用程序释放鼠标捕获
  ReleaseCapture
  '移动窗体
  SendMessage Command2.hwnd, WM_SYSCOMMAND, SC_MOVE, 0
  End If
End Sub

 command2 无法移动呢


[解决办法]
Declare Function SendMessage Lib "user32 " Alias "SendMessageA " (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, lParam As Any) As Long

多了两个空格，如下试试：
Declare Function SendMessage Lib "user32" Alias "SendMessageA" (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, lParam As Any) As Long
[解决办法]

探讨

Declare Function SendMessage Lib "user32 " Alias "SendMessageA " (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, lParam As Any) As Long

多了两个空格，如下试试：
Declare Function SendMessage Lib……

[解决办法]
Public Declare Function SendMessage Lib "user32" Alias "SendMessageA" (ByVal hwnd As Long, ByVal wMsg As Long, ByVal wParam As Long, lParam As Any) As Long