Regex allow multiple work in a sentence

Question

I'm trying to parse following sentences with regex (javascript) :

Currently I'm trying : I(\b[a-zA-Z]*\b){0,5}(TV|chocolate|fire) but it doesn't work. I also made some test with \w but no luck.

I want to allow any word (max 5 words) between "I" and the last word witch is predefined.

Wiktor Stribiżew · Accepted Answer · 2016-08-17T08:50:35.313

4

To account for non-word chars in-between words, you may use

/I(?:\W+\w+){0,5}\‌W+(?:TV|chocolate|fir‌e)/

The point is that you added word boundaries, but did not account for spaces, punctuation, etc. (all the other non-word chars) between "words".

Pattern details:

I - matches the left delimiter
(?:\W+\w+){0,5}\‌W+ - matches 0 to 5 sequences (due to the limiting quantifier {n,m}) of 1+ non-word chars (\W+) and 1+ word chars after them (\w+), and a \W+ at the end matches 1 or more non-word chars that must be present to separate the last matched word chars from the...
(?:TV|chocolate|fir‌e) - matches the trailing delimiter

edited Aug 17 '16 at 08:50

answered Aug 17 '16 at 08:27

Wiktor Stribiżew

(the *or*, using `\s` is always preferred over *spaces*) – Roko C. Buljan Aug 17 '16 at 08:28
Sometimes, a literal space is meant. BTW, here is a similar regex: [`/I(?:\W+\w+){0,5}\W+(?:TV|chocolate|fire)/`](https://regex101.com/r/iE5nA7/1) that makes sure there are non-word chars in-between the word char chunks. – Wiktor Stribiżew Aug 17 '16 at 08:31
1

The version with \W+\w+ is what I looking for ! Thanks ! :) – jaumard Aug 17 '16 at 08:47

score 0 · Answer 2 · answered Aug 17 '16 at 08:23

0

You need to add the whitespace after the I. Otherwise it wouldn´t capture the whole sentence.

I(\b[a-zA-Z ]*\b){0,5}(TV|chocolate|fire)

I greate site to test regex expressions is regexr

answered Aug 17 '16 at 08:23

Stefan Kert

It [will match `I want a b c d e f some chocolate`](https://regex101.com/r/yE9dD3/1) – Wiktor Stribiżew Aug 17 '16 at 08:25
With this expression you can ignore the mentioned `I([ ]*\b[a-zA-Z]*\b){0,5}(TV|chocolate|fire)` – Stefan Kert Aug 17 '16 at 08:27

score 0 · Answer 3 · answered Aug 17 '16 at 08:27

0

If you don't care about the spaces, use:

/I(\s[a-zA-Z]*\s?){0,5}(TV|chocolate|fire)/

answered Aug 17 '16 at 08:27

Eeko

This won't account for cases where words are separated with punctuation, that is why I suggest using `\S`/`\s`. – Wiktor Stribiżew Aug 17 '16 at 08:29

score 0 · Answer 4 · answered Aug 17 '16 at 08:40

Try

/I\s+(?:\w+\s+){0,5}(TV|chocolate|fire)/

Based on Stefan Kert version, but rely on right side spaces of each extra word instead of word boundaries.

It also accepts any valid "word" (\w) character words of any length and any valid spacing character (not caring for repetitions).

4 Answers4