Использование стандартных окон (open) в Access 2000???

можно ли использовать стандартные окна (открыть) в Access?
3 ответа

А как стандартными средствами, через API, задать директорию. т.е. выловить путь к каталогу.


Private Type BrowseInfo hWndOwner As Long pIDLRoot As Long pszDisplayName As Long lpszTitle As Long ulFlags As Long lpfnCallback As Long LParam As Long iImage As Long End TypePrivate Declare Sub CoTaskMemFree Lib 'ole32.dll' (ByVal hMem As Long) Private Declare Function SHBrowseForFolder Lib 'shell32' (lpbi As BrowseInfo) As Long Private Declare Function SHGetPathFromIDList Lib 'shell32' (ByVal pidList As Long, ByVal lpBuffer As String) As Long Private Declare Function lstrcat Lib 'kernel32' Alias 'lstrcatA' (ByVal lpString1 As String, ByVal lpString2 As String) As LongPublic Function FindFolder(strTitle As String) As String Dim iNull As Integer, lpIDList As Long, lResult As Long Dim sPath As String, udtBI As BrowseInfoWith udtBI .hWndOwner = Application.hWndAccessApp .lpszTitle = lstrcat(strTitle, '') .ulFlags = 1 End With lpIDList = SHBrowseForFolder(udtBI) If lpIDList Then sPath = String$(512, 0) SHGetPathFromIDList lpIDList, sPath CoTaskMemFree lpIDList iNull = InStr(sPath, vbNullChar) If iNull Then sPath = Left$(sPath, iNull - 1) End If End If If Len(sPath) = 0 Then Exit Function If Right(sPath, 1) <> ' ' Then sPath = sPath & ' ' FindFolder = sPath End Function


Спасибочки, все классно заработало.