Find the directory part (minus the filename) of a full path in access 97

Question

For various reasons, I'm stuck in Access 97 and need to get only the path part of a full pathname.

For example, the name

c:\whatever dir\another dir\stuff.mdb 

should become

c:\whatever dir\another dir\ 

This site has some suggestions on how to do it: http://www.ammara.com/access_image_faq/parse_path_filename.html

But they seem rather hideous. There must be a better way, right?

Answer 1

You can do something simple like: Left(path, InStrRev(path, "\"))

Example:

Function GetDirectory(path)    GetDirectory = Left(path, InStrRev(path, Application.PathSeparator)) End Function 

Answer 2

I always used the FileSystemObject for this sort of thing. Here's a little wrapper function I used. Be sure to reference the Microsoft Scripting Runtime.

Function StripFilename(sPathFile As String) As String  'given a full path and file, strip the filename off the end and return the path  Dim filesystem As New FileSystemObject  StripFilename = filesystem.GetParentFolderName(sPathFile) & "\"  Exit Function  End Function