Regex group include if condition

Question

i have try to use that regex /^(\S+)(?:\?$|$)/

with yolo and yolo?

works with both but on the second string (yolo?) the ? will be include on the capturing group (\S+).

It's a bug of regex or i have made some mistake?

edit: i don't want that the '?' included on the capturing group. Sry for my bad english.

@AvinashRaj Sry for that next time i will explain better my problem — Keltere, Jun 27 '14 at 14:02

score 4 · Accepted Answer · edited May 23 '17 at 12:11

You can use

If what you want to capture can't have a ? in it, use a negated character class [^...] (see demo here):
```
^([^\s?]+)\??$
```
If what you want to capture can have ? in it (for example, yolo?yolo? and you want yolo?yolo), you need to make your quantifier + lazy by adding ? (see demo here):
```
^(\S+?)\??$
```
There is BTW no need for a capturing group here, you can use a look ahead (?=...) instead and look at the whole match (see demo here):
```
^[^\s?]+(?=\??$)
```

What was happening

The rules are: quantifiers (like +) are greedy by default, and the regex engine will return the first match it finds.

Considers what this means here:

\S+ will first match everything in yolo?, then the engine will try to match (?:\?$|$).
\?$ fails (we're already at the end of the string, so we now try to match an empty string and there's no ? left), but $ matches.

The regex has succesfully reached its end, the engine returns the match where \S+ has matched all the string and everything is in the first capturing group.

To match what you want you have to make the quantifier lazy (+?), or prevent the character class (yeah, \S is a character class) from matching your ending delimiter ? (with [^\s?] for example).

@Massimo1993: edited answer to add possible solutions on top of what was happening :) — Robin, Jun 27 '14 at 13:51

Avinash Raj · Answer 2 · 2014-06-27T13:51:06.643

2

The below regex would capture all the non space characters followed by an option ?,

^([\S]+)\??$

DEMO

OR

^([\w]+)\??$

DEMO

If you use \S+, it matches even the ? character also. So to seperate word and non word character you could use the above regex. It would capture only the word characters and matches the optional ? which is follwed by one or more word characters.

edited Jun 27 '14 at 13:51

answered Jun 27 '14 at 13:34

Avinash Raj

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The poster does not want `?` to be in the first capture group which it is with the regex he has provided. – Mike H-R Jun 27 '14 at 13:39
@AvinashRaj look at your demo, it does what the OP doesn't want. – Mike H-R Jun 27 '14 at 13:48
@MikeH-R Did you know what the op wants? – Avinash Raj Jun 27 '14 at 13:49
@AvinashRaj I think so (correct me if I'm wrong). He is trying to make a regex that will match `yolo` and `yolo?` but will have a capture group that only says `yolo` in it. – Mike H-R Jun 27 '14 at 13:50
@MikeH-R see my second one. – Avinash Raj Jun 27 '14 at 13:51

Mike H-R · Answer 3 · 2014-06-27T13:42:50.957

2

This is the correct response as \S+ matches one or more non-whitespace characters greedily, of which ? is one.

thus the question mark is matched in the (\S+) group and the non-capturing group resolves to $ you could make it work as you expect by making the match non-greedy with:

/^(\S+?)(?:\?$|$)/

demo

alternatively you could restrict the character group:

/^([^\s?]+)(?:\?$|$)/

demo

edited Jun 27 '14 at 13:42

answered Jun 27 '14 at 13:35

Mike H-R

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Sry doesn't work. Can't recognize the end of the string. – Keltere Jun 27 '14 at 13:40

score 2 · Answer 4 · answered Jun 27 '14 at 13:47

2

Make the + non greedy:

^(\S+?)\??$

answered Jun 27 '14 at 13:47

Toto

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hwnd · Answer 5 · 2014-06-27T14:01:36.387

2

It is doing that because \S matches any non-white space character and it is being greedy.

Following the + quantifier with ? for a non-greedy match will prevent this.

^(\S+?)\??$

Or use \w here which matches any word character.

^(\w+)\??$

edited Jun 27 '14 at 14:01

answered Jun 27 '14 at 13:48

hwnd

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Regex group include if condition

5 Answers5