Match list of words without the list of chars around

Question

I have this regex

(?:$|^| )(one|common|word|or|another)(?:$|^| ) 

which matches fine unless the two words are next to each other.

One one's more word'word common word or another word more another   More and more years to match one or more other strings  And common word things and or 

In the above it matches one in line two but not the or just next to it. Same for common and word int the third line.

Live Example: http://regex101.com/r/hV3wQ3

I believe it's something to do with the non-matching groups' number. But, I am not sure how to achieve the end goal of matching all the list of words without any char around them.

I do not want the one in one's or the word in word'word to be matched.

Answer 1

Since your capture groups define explicitly one character on either side of the common word, it's looking for space word space and then when it doesn't find another space, it fails.

In this case, since you don't want to match all the characters word boundary's would catch (period, apostrophe, etc.) you need to use a bit of trickery with lookaheads, lookbehinds, and non-capture groups. Try this:

(?:^|(?<= ))(one|common|word|or|another)(?:(?= )|$) 

http://regex101.com/r/cM9hD8

Word boundaries are still simpler to implement, so for reference sake, you could also do this (though it would include ', ., etc.).

\b(one|common|word|or|another)\b 

Answer 2

You can use (?:[\s]|^)(one|common|word|or|another)(?=[\s]|$) instead.

It will not match one's , someone ,etc...

Check DEMO