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!

RegEx.yxmd

Workflow

Transformation

Accepted answers

ShankerV

Hi @User12

Please find the expected output via Regex as expected.

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

Breaking down the Regex formula built to arrive the solution,

we have a unique pattern in the Input.

Hence .*\\ will help to eliminate the characters in the front.

The regex tool will create a new column only for the data inside the () open and close paranthesis

Hence I am reading (.*) where it will read all the characters . denotes any charater and * denotes n number

.\l denotes small a-z and when it finds .xlsx it stops reading the filename.

hope this helps!!!!

binu_acs

@User12 Attaching the workflow

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

All comments

ShankerV

hi @User12

Please let me know what is the expected output.

I see only the input in the attached workflow. Sharing the expected output will help to work on the solution.

binu_acs

@User12 One way of doing this with the Regex_Replace formula

Regex_1213.yxmd

User12

Hi!

I hope to get the next output:

18.7.

18.8+

18.9_

ShankerV

@User12

Does filename you are expecting is "18.7..xlsx" or "18.7."

JamesCharnley

Hi @User12,

FileGetFileName returns the name portion of the file without the extension. It's essentially taking everything from the last '\' to the next '.', signifying the start of the extension. When I use it on your input, it's working as expected with the output that you want:

Is there some reason this doesn't work for your file names? It should be pretty robust

FileGetFileName.yxmd

binu_acs

@User12 if you don't want the extension then update the formula like below

User12

That's exactly what I need! Thank you!

But, could you, please, provide me with the expression (I don't really understand exact signs that you are using), that i can copy&paste into my formula tool?
It would be very helpful.

binu_acs

@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.

Isolate filename from full path_Non-RegEx.yxmd

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