Hello,
I receive a list of FullPaths and i need to extract the FileName.
If anyone could help with a way to parse the name i will appreciate it (The last "\" will be the delimiter).
Solved! Go to Solution.
You dont need regex you can directly use formula FileGetFileName([Field1])
Hope this helps : )
Try this function in a Formula tool:
FileGetFileName(Path): Returns the name portion of the path, without the extension.
List of all functions: https://help.alteryx.com/20213/designer/functions-quick-reference
Chris