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