What does this notation mean "?:" in regex?

Question

I found the below regex from RegexLib.

^(?:(?:(?:0?[13578]|1[02])(\/|-|\.)31)\1|(?:(?:0?[13-9]|1[0-2])(\/|-|\.)(?:29|30)\2))(?:
(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:0?2(\/|-|\.)29\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468]
[048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:(?:0?[1-9])|(?:1[0-2]))
(\/|-|\.)(?:0?[1-9]|1\d|2[0-8])\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$

It is from http://regexlib.com/REDetails.aspx?regexp_id=113

I don't understand the notation "?:", which is used many times in this regex. It seems ":" is not in the chartsheet

I am that familiar with regex. Could anyone show an example, with the regex and inputs?

For some extra explanation check [this](http://stackoverflow.com/q/3512471/557306) question/ — sidyll, Mar 12 '13 at 21:13

stema · Answer 1 · 2013-03-12T21:43:56.223

7

(?:...) is just a non capturing group, means the part of the string that is matched by this group is not stored and can not be accessed by $1 or \1

Capturing groups are numbered by the opening brackets, so

For the example text "Foobar test"

Here are two capturing groups

^(\w+)\s*(\w+)
 1       2

This will result in:

$1 = "Foobar"

$2 = "test"

Here is only one capturing group:
```
^(?:\w+)\s*(\w+)
           1
```
This will result in:
```
$1 = "test" 
```

edited Mar 12 '13 at 21:43

answered Mar 12 '13 at 21:11

stema

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@JackWM, an example of what? sidyll's link has a good explanation with example. – stema Mar 12 '13 at 21:21
@JackWM, I added an example. – stema Mar 12 '13 at 21:37

score 2 · Answer 2 · answered Mar 12 '13 at 21:12

2

(?:...) is a non-capturing group, that is, it controls operator precedence but it won't create a match group.

answered Mar 12 '13 at 21:12

Joey

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score 2 · Accepted Answer · edited Jun 20 '20 at 09:12

2

(?:)

stands for non capturing group. See perldoc perlre (perl is IMHO the most advanced language to play with REGEX : pcre by example, means Perl Compatible Regular Expression and is the default for PHP, pcregrep or with grep -P)

Relevant part of perl doc :

(?:pattern)
(?adluimsx-imsx:pattern)
(?^aluimsx:pattern)

This is for clustering, not capturing; it groups subexpressions like "()", but doesn't make backreferences as "()" does.

See this example using perl :

$ echo 'azfoobar' | perl -lne 'print $1 if /^(?:az)(.*)/'
foobar

As you can see, the first matching capturing group ($1) isn't az but the rest of the line. This works the same with the others languages like python or such.

edited Jun 20 '20 at 09:12

Community

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answered Mar 12 '13 at 21:23

Gilles Quénot

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Thanks. I just found a similar notation `(?i:...)` in `(?i:(?=[MDCLXVI])((M{0,3})((C[DM])|(D?C{0,3}))?((X[LC])|(L?XX{0,2})|L)?((I[VX])|(V?(II{0,2}))|V)?))`. Any thoughts? – JackWM Mar 12 '13 at 21:29
1

@JackWM, `(?i:...)` is a combination of the inline modifier `i` that turns on case independent matching and a non capturing group. Means: The pattern inside this group is not case sensitive and is not captured. – stema Mar 12 '13 at 21:36
Added explanations and perl doc – Gilles Quénot Mar 12 '13 at 21:45
1

PCRE is not a standard, it's the name of a library ([ref](http://www.pcre.org/pcre.txt)). There **is** no standard for regexes. Case in point: the `(?^...)` construct is new in Perl v5.14. I had never heard of it until now, and I would be extremely surprised if any other flavor has picked it up yet. – Alan Moore Mar 13 '13 at 04:44

What does this notation mean "?:" in regex?

3 Answers3