I want to know how can we open a file dialog from withing a class module in vb6. I know how to do in forms, but I have to open it from within a class module.
+1
A:
Have a look at the API below:
Private Declare Function GetOpenFileName Lib "comdlg32.dll" Alias _
"GetOpenFileNameA" (pOpenfilename As OPENFILENAME) As Long
Here's a sample of it in use:
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
Private Function FileOpenDialog()
Dim sFilter As String
OpenFile.lStructSize = Len(OpenFile)
sFilter = "Text Files (*.txt)" & Chr(0) & "*.TXT" & 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 = "C:\"
OpenFile.lpstrTitle = "Select File"
OpenFile.flags = 0
lReturn = GetOpenFileName(OpenFile)
End Function
pm_2
2010-05-26 07:26:16
+1
A:
Yes, you can call the API to raise this dialog.
Most of the time though this sort of need stems from a broken paradigm. A Class should not have a UI. When you have a real need for this your Class should probably be a UserControl... and the problem goes away.
Bob Riemersma
2010-05-27 02:49:30