String Parsing

I have a text column with following information. I want to extract the workflow name from the column.

Started running \ProgramData\Alteryx\Service\Staging\34ab1bb62d8c23eaae8ae\test.yxmd at Wed Sep 6 05:00:43 2023
Started running \ProgramData\Alteryx\Service\Staging\34ab1bb62d23eaae8ae\test1.yxmd at Thurs Sep 7 04:00:43 2023
Started running \ProgramData\Alteryx\Service\Staging\34ab1bb62d8c23eaae8ae\test2.yxmd at Tues Sep 4 07:00:43 2023
Started running \ProgramData\Alteryx\Service\Staging\34ab1bb62d8c28ae\test3.yxmd  at Thurs Aug 7 04:00:43 2023
Started running \ProgramData\Alteryx\Service\Staging\34ab1bb6aae8ae\test4.yxmd  at Thurs July 7 06:00:43 2023
Started running \ProgramData\Alteryx\Service\Staging\34ab1bb62d8c23eaa\test5.yxmd  at Thurs Aug 7 07:00:43 2023

Desired output

test.yxmd

test1.yxmd

test2.yxmd

test3.yxmd

test4.yxmd

test5.yxmd

Appr

Regex

Accepted answers

All comments

khanp27

Input information is provided above

alexnajm

This formula in a Formula tool should do the trick: Right([Column],FindString(ReverseString([Column]),"\"))

binu_acs

@khanp27

Regex_Replace([input],”.*\\(.*)”,”$1”)

HarperJ

Assuming that's all in one field, the formula should be: right(getword([Text],2),STRCSPN(ReverseString(getword([Text],2)),'\'))

alexnajm

It looks like the requirements changed from the initial post - what are the final values?

binu_acs

@khanp27 One way of doing this

Regex_070923.yxmd

flying008

Hi, @khanp27

FYI.

REGEX_Replace([Txt], '.+\\([^\\]+.yxmd)(?=\s).*$', '$1')

Rags1982

If you want the file name without the extension, you can just use the formula FileGetFileName([field]).

caltang

@khanp27 Try this REGEX Parse:

.*\\(.*?\.yxmd)

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