Regex only capture first match

Question

I want to parse a string using regex, example of the string is

Lot: He said: Thou shalt not pass!

I want to capture Lot as a group, and He said: Thou shalt not pass!. However, when I used my (.+): (.+) pattern, it returns

Lot: He said: and Thou shalt not pass!

Is it possible to capture He said: Thou shalt not pass using regex?

do you want to capture the second part without `!`? – Avinash Raj Sep 01 '14 at 09:44 — Avinash Raj, Sep 01 '14 at 09:44

score 41 · Accepted Answer · answered Sep 01 '14 at 09:36

41

You need a non-greedy (or lazy) cardinality for the first group: (.+?): (.+).

More details on http://www.regular-expressions.info/repeat.html, chapter "Laziness Instead of Greediness".

answered Sep 01 '14 at 09:36

sp00m

Works like a charm, @sp00m sorry for not able to upvote because my lack of reps. – Wancak Terakhir Sep 01 '14 at 11:53

score 5 · Answer 2 · answered Sep 01 '14 at 09:36

5

give this a try:

([^:]+):\s*(.*)

answered Sep 01 '14 at 09:36

Kent

+1 for using negative character class rather than `.` – asontu Sep 01 '14 at 09:38
@funkwurm negative character class is considered faster than non-greedy matching. – Kent Sep 01 '14 at 09:43

score 2 · Answer 3 · answered Sep 01 '14 at 09:37

2

(?=.*?:.*?:)(.*?):(.*)

You can use this.

See demo.

answered Sep 01 '14 at 09:37

vks

score 2 · Answer 4 · answered Sep 25 '17 at 09:21

2

You should use the flags ig - i = case insensitive, g = don't return after first match and a expression like "(.:) ?(.: .*)".

answered Sep 25 '17 at 09:21

Matheus Dadalto

score 0 · Answer 5 · answered Sep 01 '14 at 09:44

0

The following one works.

([^:]+): (.+)

answered Sep 01 '14 at 09:44

Venky

Could you please explain it in more details? – Alex Netkachov Sep 01 '14 at 10:08
please see the following link for explanation [link](http://regex101.com/r/jE3jU8/1) – Venky Sep 01 '14 at 10:58

5 Answers5