Strange regex exception with fullstop [duplicate]

Question

I am using

^(?i)(?=.*\bWORD_TO_FIND\b).*$

In this specific case I am trying to match "S.A.M."

This way it works:

^(?i)(?=.*\bs.a.m\b).*$

This way it doesn't:

^(?i)(?=.*\bs.a.m.\b).*$

Why does that fullstop brakes the match?

You don't need the lookahead assertion for this. And you should regex quote the find word(s). And you should use a boundary with qualification. You could use a conditional boundary "(?i)^.*(?(?=\\w)\\b)(?:" + regexQuote(text) + ")(?(?<=\\w)\\b).*$" or, the best way is to use a whitespace boundary "(?i)^.*(?<!\\S)(?:" + regexQuote(text) + ")(?!\\S).*$" — user557597
– user557597, Commented Nov 18, 2015 at 17:10

vks · Accepted Answer · 2015-11-18 14:56:14Z

1

That is because of \b or word boundary.After . there is no \b or word boundary but after m there is.

^(?i)(?=.*\bs\.a\.m\.\b).*$ ^^ Here no word boundary so assertion or lookahead fails.

You should also be escaping . if you want to match .

answered Nov 18, 2015 at 14:56

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5 Comments

Thanks, what other symbols should I escape?

@user2468425 all metacharacters that have special meaning in regex like .*?{} etc.

@user2468425 do accept if it worked for you

I'm sorry but it doesn't work for me

@user2468425 what doesnt work?see here regex101.com/r/hE4jH0/13 for java you need to use ^(?i)(?=.*\\bs\\.a\\.m\\b).*$