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
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