Today is my first day learning about the Regular expressions and I have accepted defeat.
The only thing I need it for is one specific thing, and I do not want to spend a weekend studying regular expressions.
Here is a doc that shows what I want to achieve.
Here you go @Fran_Vidicek 
RegexReplace(thisRow.Column,"r\d{1,2}","p0" )
The Regular Expression "r\d{1,2}" will match the character r and any digit after that (but for a count in between 1 and 2 (\d{1,2}) and replace them by p0
Thank you very much!
It seems like there is someone active in the community at any hour of the day, that’s nice.
What if there is "r" with more than 2 numbers after it (e.i. r463718/r3/4/r5)?
Is there a way to get the same result no matter the amount of numbers after the "r"
Is there something more elegant than ("r\d{1,9999}","p0" ) ?
Or would that even work the way I think it would.
No problem @Fran_Vidicek 
!
I’m honestly not a RegEx expert but this seems to work (I’ve added it to the sample above too )
RegexReplace(thisRow.Column,"r\d+","p0" )
The + means « one or more » so, \d+ is one or more digits
(And lol, yes, there’s always someone here 
… I should be sleeping by now but here I am 
)
Sweet! Thank you once again 
Yeah, me too. It’s 3:00 AM here.
You could do something a bit simpler in this case without regex!
You could likely just do
Repeat(“p0/“, column.split(“/“).count())
All that is doing is counting how many “r/“ things you have regardless of the number of digits after the r and then repeating your p0 string a matching number of times
Oh, thank you, that could come in handy too.
Could this function possibly be slightly faster than regex, when it comes to executing it ?
I know that feeling when you’re kind of stuck in your coda doc
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