Extracting file name from customized path

[User12]

Hello everyone!

I have a list of paths that lead to exact sheet, but I need only File name.

I suppose it should be a solution via RegEx ( I don't know how to use this tool properly)

Formula tool is not working well in my example (Filegetfilename).

Could you, please, provide me with the suitable expression? 

Workflow with the data attached below.

Thanks in advance!

[binuacs · Accepted solution]

@User12 Attaching the workflow

REGEX_Replace([Path], '.*\\(.*)\.xlsx.*', '$1')

[ShankerV]

Hi @User12 

Regex toll has 4 different output methods. You can use which suits your need, I have used Parse as I need to extract the output to a new column.

1. Parse helps to create a new column for the output.

Each () will create a new output.

.*\\(.*)(.\l{4})\|\|\|.*

Below cheat sheet will help you, 

Let me know if you have more questions.

Many thanks

Shanker V

[PhilipL]

@User12  If you don't want to use RegEx, here's a solution that takes a different path. :)

[User12]

Thank you so much! It's really helpful :)