Get the Full Path of a File

This is a very simple way to get just the full directory path of a file passed in.

Public Function GetTheFilePath(ByVal strFile) As String
   Files = Split(strFile, "\")
   For I = 0 to (UBound(Files) - 1)
      FilePath = FilePath & Files(I) & "\"
   Next
   GetTheFilePath = FilePath
End Function

Usage

MyFilePath = GetTheFilePath("C:\testing\jim\file.txt")

will return:

C:\testing\jim\