Extract filename from path

Question

I need to extract the filename from a path (a string):

e.g., \"C:\\folder\\folder\\folder\\file.txt\" = \"file\" (or even \"file.txt\" to get me sta

Answer 1

Here's simpler method: a one-line function to extract only the name — without the file extension — as you specified in your example:

Function getName(pf):getName=Split(Mid(pf,InStrRev(pf,"\")+1),".")(0):End Function

...so, using your example, this:

       MsgBox getName("C:\folder\folder\folder\file.txt")

  returns:

       

For cases where you want to extract the filename while retaining the file extension, or if you want to extract the only the path, here are two more single-line functions:

Extract Filename from x:\path\filename:

Function getFName(pf)As String:getFName=Mid(pf,InStrRev(pf,"\")+1):End Function

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Extract Path from x:\path\filename:

Function getPath(pf)As String: getPath=Left(pf,InStrRev(pf,"\")): End Function

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Examples:

^(Source)