Could someone explain the regex /(.)\.(.)/?

Question

I want to get the file extension in Groovy with a regex, for let's say South.6987556.Input.csv.cop.

http://www.regexplanet.com/advanced/java/index.html shows me that the second group would really contain the cop extension. Which is what I want.

0: [0,27] South.6987556.Input.csv.cop
1: [0,23] South.6987556.Input.csv
2: [24,27] cop

I just don't understand why the result won't be

0: [0,27] South.6987556.Input.csv.cop
1: [0,23] South
2: [24,27] 6987556.Input.csv.cop

What should be the regex to get this kind of result?

The reason is that the (.*) is "greedy" - it will gobble up as many characters as possible. To make it not-greedy, add a question mark. `(.*?)\.(.*)`. — Alec, Jul 28 '14 at 06:08
Does the last `?` belongs to the regex or not? Why don't you play with [regex101](http://regex101.com/r/sQ0kW4/1)? — Avinash Raj, Jul 28 '14 at 06:09
Please see http://stackoverflow.com/questions/22937618/reference-what-does-this-regex-mean ... [You also probably want `(.*?)\.(.*)`](http://stackoverflow.com/questions/24989252/could-someone-explain-reg-exp/24989310#comment38850267_24989252) — Unihedron, Jul 28 '14 at 06:11
@AvinashRaj: no, ? is not part of the reg exp. I updated the title — Radek, Jul 28 '14 at 06:13
@Alec: could you make an answer from your comment so I can except it? — Radek, Jul 28 '14 at 06:16

score 2 · Answer 1 · edited Jul 28 '14 at 07:17

To get the desired output, your regex should be:

((.*?)\.(.*))

DEMO

See the captured groups at right bottom of the DEMO site.

Explanation:

(         group and capture to \1:
  (       group and capture to \2:
    .*?   any character except \n (0 or more
          times) ? after *  makes the regex engine
          to does a non-greedy match(shortest possible match).
  )       end of \2
  \.      '.'
  (       group and capture to \3:
    .*    any character except \n (0 or more
          times)
  )       end of \3
)         end of \1

bukart · Accepted Answer · 2014-07-28T06:25:59.463

2

Here is a visualization of this regex

(.*)\.(.*)

Regular expression visualization

Debuggex Demo

in words

(.*) matches anything als large as possible and references it
\. matches one period, no reference (no brackets)
(.*) matches anything again, may be empty, and references it

in your case this is

(.*) : South.6987556.Input.csv
\. : .
(.*) : cop

it isn't just only South and 6987556.Input.csv.cop because the first part (.*) isn't optional but greedy and must be followed by a period, so the engine tries to match the largest possible string.

Your intended result would be created by this regex: (.*?)\.(.*). The ? after a quantifier (in this case *) switches the behaviour of the engine to ungreedy, so the smallest matching string will be searched. By default most regex engines are greedy.

edited Jul 28 '14 at 06:25

answered Jul 28 '14 at 06:18

bukart

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no, it still allows, like before, to have a filename with en empty extension – bukart Jul 28 '14 at 06:20
argh... right... I forgot the `*` and thought about `+`... sry... you're right – bukart Jul 28 '14 at 06:22
@bukart it won't capture the entire string. – Avinash Raj Jul 28 '14 at 06:25
updated my answere corresponding to the updated regex in the title – bukart Jul 28 '14 at 06:26
I was mainly about the explanation of the desired and working well reg exp `(.*)\.(.*)`. Thank you. – Radek Jul 28 '14 at 06:30
@bukart : **question :** why doesn't `\.` matches `.` after `South` but only before `cop`? – NoobEditor Jul 28 '14 at 06:53
@NoobEditor: because the first group (.*) catches as many characters as possible to the last dot. That's what I understood. :-) – Radek Jul 28 '14 at 07:30

Could someone explain the regex /(.*)\.(.*)/?

2 Answers2

Could someone explain the regex /(.)\.(.)/?