Открытие файла по умолчанию - MS Access - Ответ 2681980

Private Declare Function GetOpenFileName Lib "comdlg32.dll" Alias _
    "GetOpenFileNameA" (pOpenfilename As OPENFILENAME) As Long
 
Private Type OPENFILENAME
    lStructSize As Long
    hwndOwner As Long
    hInstance As Long
    lpstrFilter As String
    lpstrCustomFilter As String
    nMaxCustFilter As Long
    nFilterIndex As Long
    lpstrFile As String
    nMaxFile As Long
    lpstrFileTitle As String
    nMaxFileTitle As Long
    lpstrInitialDir As String
    lpstrTitle As String
    flags As Long
    nFileOffset As Integer
    nFileExtension As Integer
    lpstrDefExt As String
    lCustData As Long
    lpfnHook As Long
    lpTemplateName As String
End Type
 
'    MsgBox OpenFileDlg("Выберите текстовый файл", "C:"', "Текстовые документы (*.txt)", "*.txt")
Public Function OpenFileDlg(Title, InitDir, FilterName, Mask As String) As String
    Dim OpenFile As OPENFILENAME
    Dim lReturn As Long
    Dim sFilter As String
    Dim i As Integer
    OpenFile.lStructSize = Len(OpenFile)
    OpenFile.hwndOwner = Application.hWndAccessApp
    sFilter = FilterName & Chr(0) & Mask & Chr(0)
    OpenFile.lpstrFilter = sFilter
    OpenFile.nFilterIndex = 1
    OpenFile.lpstrFile = String(257, 0)
    OpenFile.nMaxFile = Len(OpenFile.lpstrFile) - 1
    OpenFile.lpstrFileTitle = OpenFile.lpstrFile
    OpenFile.nMaxFileTitle = OpenFile.nMaxFile
    OpenFile.lpstrInitialDir = InitDir
    OpenFile.lpstrTitle = Title
    OpenFile.flags = 0
    lReturn = GetOpenFileName(OpenFile)
 
    If lReturn = 0 Then
        OpenFileDlg = ""
    Else
        i = 1
        While (Asc(Mid(OpenFile.lpstrFile, i, 1)) <> 0)
            i = i + 1
        Wend
        OpenFileDlg = Left(OpenFile.lpstrFile, i - 1)
    End If
End Function