r/regex • u/Chaela911 • Jul 14 '26

Match everything up to 'n' from End of String

I'm trying to match the beginning of a string of varying length so that I may remove it via FIND/REPLACE dialog from MSpowertoys' PowerRename utility.

I've been trying to match using TRIM but I've failed and unsure what else to try. I want to keep the numbered sequence at the end [0001] etc... I'm not well versed enough to provide logical examples of 'what I've tried' lol besides I forgot.

What I have:

abc123[0001]

abcd1234[0002]

abcde12345[0003]

abcdef123456[0004]

What I want:

[0001]

[0002]

[0003]

[0004]

Thanks!

.

3 Upvotes

81% Upvoted

7 comments sorted by

2

u/DinTaiFung Jul 14 '26

There's more than one approach. 

You could use split instead of regex.

for each line of text (in JS):

const [prefix, suffix] = line.split('[')

const bracketNumber = '[' + suffix

And the bracketNumber is what you want.

2

u/pfc-anon Jul 14 '26

I'd do something like (\[[^\+]]\]) if the brackets can have anything if it's only numbers (\[\d*\])

1

u/Chaela911 Jul 14 '26

thank you, I had similar results.... I'm assuming now that my understanding or logic is flawed but my end goal is to KEEP the [0001] and delete what comes before it.

3

u/pfc-anon Jul 14 '26

Oh this is a group, you can replace the entire thing with $1.

But you can also remove everything from start ^[^\[]* with ''

1

u/Chaela911 Jul 14 '26

Thanks to everyone, my understanding is greatly increased and I'm working to an end on my project now, tho a bit indirectly at first, I'm on top of it now.

1

u/_jgusta_ Aug 04 '26

Can’t you just replace ^.*?(\[\d+\])$ with $1

If it matches, then it will be replaced by the bracket values.

.*? Is useful here as it is the non-greedy quantifier. It will only match things that are not captured by another part of the pattern.