Regex to not match a pattern in string

Question

I am a newbie and have been struggling the last hour to figure this out. Let's say you have these strings:

baa cec haw heef baas bat jackaay

I want to match all the words which don't have two aa's consecutively, so in the above it will match cec, haw, heef, bat.

This is what i have done so far, but it's completely wrong i can sense :D

\w*[^\s]*[^a\s]{2}[^\s]*\w*

@Kevin Guan: Changed just a sec before you posted the comment, :p. He said, `strings` which I interpreted _a collection/list_ — mshsayem, Nov 07 '15 at 11:28
@mshsayem I disagree, use regex on a *collection/list* without `for` loop? whose idea? :P — Remi Guan, Nov 07 '15 at 11:30
Related: [Regular Expression block negation](http://stackoverflow.com/q/14819125/319403) — cHao, Nov 07 '15 at 11:52
Where are these strings held? In a file, a space delimited string or something else? — Martin Evans, Nov 07 '15 at 13:10
@StefanBratanov: Regrettably, differences among regex flavors can't be ignored. even for a question as simple as this one. And using multiple flavor tags just confuses things; I've removed them. — Alan Moore, Nov 07 '15 at 13:40

score 1 · Answer 1 · edited May 23 '17 at 10:27

1

You maybe want to use negative lookahead:

/(^|\s)(?!\w*aa\w*)(\w+)/gi

You can check your string by paste this code on console on Chrome/Firefox (F12):

var pattern = /(^|\s)(?!\w*aa\w*)(\w+)/gi;
var str = 'baa cec haw heef baas bat jackaay';
while(match = pattern.exec(str))
    console.log(match[2]); // position 2 is (\w+) in regex

You can read more about lookahead here. See it on Regex101 to see how this regex work.

edited May 23 '17 at 10:27

Community

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answered Nov 07 '15 at 12:32

Van M. Tran

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Thank you. This almost works. It just captures the spaces between words as well as part of the group. – Stefan Bratanov Nov 07 '15 at 13:55
If you select 2nd group, you will get expected string. But I think @stribizhev has better solution. – Van M. Tran Nov 07 '15 at 17:18

Wiktor Stribiżew · Answer 2 · 2015-11-07T16:01:10.193

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You need a regex that has 2 things: a word boundary \b and a negative lookahead right after it (it will be sort of anchored that way) that will lay restrictions to the subpattern that follows.

\b(?!\w*aa)\w+

See the regex demo

Regex breakdown:

\b - word boundary
(?!\w*aa) - the negative lookahead that will cancel a match if the word has 0 or more word characters followed by two as
\w+ - 1 or more word characters.

Code demo:

var re = /\b(?!\w*aa)\w+/gi; 
var str = 'baa cec haw heef bAas bat jackaay bar ha aa lar';
var res = str.match(re);
document.write(JSON.stringify(res));

edited Nov 07 '15 at 16:01

answered Nov 07 '15 at 15:10

Wiktor Stribiżew

607,720
39
448
563

Thank you so much. I will try hard to understand how this works, but it works and is exactly what i wanted to achieve. – Stefan Bratanov Nov 07 '15 at 17:07
Please see [*Word Boundaries*](http://www.regular-expressions.info/wordboundaries.html) and [*Lookahead and Lookbehind Zero-Length Assertions*](http://www.regular-expressions.info/lookaround.html) articles. The main pattern is `\w`(alphanumeric+underscore matching subpattern) that we match 1 or more times (`+`), only after a non-word character (not `[a-zA-Z0-9_]`) if it has no `aa` (since we check the word first with the lookahead if it has `aa` after zero or more word caracters (`\w*`)). – Wiktor Stribiżew Nov 07 '15 at 17:13

score 0 · Answer 3 · answered Nov 07 '15 at 11:47

in javascript, you could use filter and regex invert ! a non-capturing group ?:.

var strings = ['baa','cec','haw','heef','baas','bat','jackaay'];
strings = $(strings).filter(function(index, element){
   return !/.*(?:aa).*/.test(element);                // regex => .*(?:aa).*
});

Regex to not match a pattern in string

3 Answers3