Function reference
Returns a named part of the file name or path.
Use this function to extract for example the extension or a particular folder level name

GetFileNamePart( ‘filename’ , ‘Name/NameNoExt/Ext/Folder-1/Folder-2/Folder-..’)

Argument Description
File name The name to process.
Option Name/NameNoExt/Ext/Folder-1/Folder-2/Folder-…

Data type: String
Returns the fraction of the file name as text.

Example 1
GetFileNamePart( ‘f:\myfolder\myfile.txt’ , ‘Ext’ )

The above example returns ‘txt’

Last modified: 29 June 2023