Max Wanadoo
max.wanadoo at gmail.com
Thu Jan 7 11:34:27 CST 2010
Why not something simple like this.. Function x() ' the following two examples will both give the same result. Dim str As String str = "c:\myfile.txt" str = Mid(str, InStrRev(str, "\") + 1) Debug.Print str str = "myfile.txt" str = Mid(str, InStrRev(str, "\") + 1) Debug.Print str End Function So, all you need is str = Mid(str, InStrRev(str, "\") + 1) Max -----Original Message----- From: accessd-bounces at databaseadvisors.com [mailto:accessd-bounces at databaseadvisors.com] On Behalf Of Heenan, Lambert Sent: 07 January 2010 17:21 To: 'Access Developers discussion and problem solving' Subject: Re: [AccessD] Common Dialog Here's a simple function to return the file path... Function GetPath(aPath) As String ' Strips the path name from the supplied file and path name ' leaves the trailing slash on there Dim foo As Integer, aSlash As Integer aSlash = 0 foo = InStr(aPath, "\") While (foo > 0) aSlash = foo foo = InStr(aSlash + 1, aPath, "\") Wend If aSlash > 0 Then GetPath = left$(aPath, aSlash) Else GetPath = "" End If End Function And using that you can get the file name... Function GetFileName(aPath) As String Dim fPath As String fPath = GetPath(aPath) If Len(fPath) = Len(aPath) Then ' only a path was provided GetFileName = "" Else GetFileName = right$(aPath, Len(aPath) - Len(fPath)) End If End Function These two functions were written back in Access 97 days, before the Split() function came to be. With split you could do something like this (air code)... Function GetFileName(sPath as String) as String vArray as Variant vArray = Split(sPath,"\") GetFileName = Cstr(Ubound(vArray)) End Function Lambert -----Original Message----- From: accessd-bounces at databaseadvisors.com [mailto:accessd-bounces at databaseadvisors.com] On Behalf Of Lawrence Mrazek Sent: Thursday, January 07, 2010 12:00 PM To: Access Developers discussion and problem solving Subject: [AccessD] Common Dialog Hi Folks: I'm the API from http://www.mvps.org/access/api/api0001.htm, and can't seem to remember how to ONLY retrieve the filename (not the filename and path using the API. I suppose that if I know the directory, I can use some functions to trim off the directory path, but in this case I might not always be looking for the files in a default directory. Any hints/guidance would be appreciated. Larry Mrazek lmrazek at lcm-res.com 314-432-5886 -- AccessD mailing list AccessD at databaseadvisors.com http://databaseadvisors.com/mailman/listinfo/accessd Website: http://www.databaseadvisors.com -- AccessD mailing list AccessD at databaseadvisors.com http://databaseadvisors.com/mailman/listinfo/accessd Website: http://www.databaseadvisors.com