Here Is My Regex Code:
/fun(niest|!ny$)?/ig
How would I get the word "fun" or "funniest" but not the word "funny" through regex, here is what I have. Is there any way of doing this, if so please help!
1 Answer
You can use word boundaries \b and an optional group (?:niest)?:
/\bfun(?:niest)?\b/ig
See the regex demo
The pattern matches:
\b- leading word boundaryfun- literal character sequencefun(?:niest)?- an optional (one or zero occurrences)niestliteral character sequence (not captured into any group since the group is non-capturing, i.e. used only for grouping)\b- trailing word boundary.
Your fun(niest|!ny$)? matches fun, or funniest or fun!ny that is at the end of the string.
answered Apr 22, 2016 at 20:31
Wiktor Stribiżew
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6 Comments
antipanchi
how would this be done with multiple options in the regex which are separated with the |, not as an option phrase/letter but as a word like funny the honey e.g: funny|honey
Wiktor Stribiżew
funny|honey will match either funny or honey. Could you be more precise?Wiktor Stribiżew
You thought that to match a whole word you need to include a "negative alternative"? That is possible in PHP (PCRE), not in JS. Either use
\b (as I showed) or non-ambiguous /(^|\W)(fun(?:niest)?)(?=\W|$)/ig (but then you will need to use backreferences wisely if replacing, or do more post-processing if matching).
Timothy Kanski
Another option: (funniest|fun(?!ny))
Wiktor Stribiżew
@TimothyKanski: If you insist, then again, it should be
/\b(funniest|fun(?!ny\b)\w*)\b/ig - otherwise I see no use for the lookahead. The thing is, the words are to be matched as whole words in the current scenario, and the 2 words are to be matched only. If OP says any word having or starting with fun should be matched, but not funny, then the lookahead solution will be the best. |
/fun(?!ny)(niest)?/