Regex with remove before and first after

Question

I have a simple example as follows

user=djdjdjjdjdjdj; user=jdjdjdjdjdj;

And

user=djdjdjjdjdjdj;user=jdjdjdjdjdj;

Differences in white space between the two examples above. When I used Regex to separate user codes, it ok with first example. Can you advise me how to do better, this is my Regex code:

(?<=user=)([^\s]+)(?=;)

What is your matching criteria? What do you need as your result? — PM 77-1, Jul 12 '20 at 02:01
Look at first example that is correct. But the data entered must have spaces, I want if no spaces remain the same. — Cosi Nguyen, Jul 12 '20 at 02:08
It's still not clear. Have a look: https://regex101.com/r/TC0GMU/1 Is it what you were asking about? — PM 77-1, Jul 12 '20 at 02:15

score 2 · Answer 1 · answered Jul 12 '20 at 02:20

I would just use here:

(?<=user=)[^;]+

Demo

This would match each user value, sans semi colon, which appears in the input. If you really want to use capture groups, then I suggest not even using the lookbehind:

\buser=([^;]+)

This would place each user value in the first capture group.

Seblor · Accepted Answer · 2020-07-12T02:27:21.007

0

If you want to match [^\s] but also a space, simply use ..

So your regex would be (?<=user=)(.+?)(?=;) (with a non greedy quantifier to match until the next character is a ;).
Or you can match until the first ;, use this : (?<=user=)([^;]+)

edited Jul 12 '20 at 02:27

answered Jul 12 '20 at 02:16

Seblor

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Regex with remove before and first after

2 Answers2

Demo