File Information

Help!

I need to find a way to obtain a file name given the path to the file.

For example if the path to the file is
D:\blah\foo\bar\file.fff

then i need to return "file.fff"
I also need to be able to do this with URL's so
http://blah/foo/bar/file.fff
will also return "file.fff"

Thanks!

Shoud work with both

Code:

---

Function Filepart(filepath)
    For n = Len(filepath) To 1 Step -1
        If Mid(filepath, n, 1) = "\" Then Exit For
    Next n
    Filepart = Mid(filepath, n + 1)
End Function

---

thanks for the help. i added a clause to handle the fact that url's use forward slash instead of backslash. otherwise it works perfectly.