Why is my regex putting a character in a different capture group than what I am expecting?

Question

I have the given strings:

bar"{foo}"bar
bar{foo}bar

I am trying to get both the first and third capture group to be just bar. The second capture group should be {foo} both with and without quotes for each match respectively.

I have the following regex:

(^.*)("?\{.*\}"?)(.*$)

With these results:

Match 1
Full match.  bar"{foo}"bar
Group 1.     bar"
Group 2.     {foo}"
Group 3.     bar

Match 2
Full match.  bar{foo}bar
Group 1.     bar
Group 2.     {foo}
Group 3.     bar

Why are the " characters not both in the second group? I do not understand why it would be in the first if I am specifically calling it out in the second group. Do I need to tell the first group to ignore it or use it as a right bound?

I am on a phone, so I give just some hints: in Match1 Group1 you use .* which consumes the " character (. means any) , so it gets matched, because following "? matches zero or one ", thus matching 0, because " was consumed by first capturing group. Try debugging pattern on regex101 — Michał Turczyn, Apr 01 '20 at 15:36
@MichałTurczyn Thank, you, you are correct. I realized my issue upon posting. — notsodev, Apr 09 '20 at 14:27

score 0 · Accepted Answer · answered Apr 01 '20 at 15:26

As I was finishing writing this question I realized the problem I had. I needed to use a reluctant quantifier so the first group would not grab the " before the next group had a shot.

Here is the small correction - I added .*? (reluctant) instead of .* (greedy):

(^.*?)("?\{.*\}"?)(.*$)

Good explanation here: Greedy vs. Reluctant vs. Possessive Quantifiers

Why is my regex putting a character in a different capture group than what I am expecting?

1 Answers1