How to find group of words that is in random order(only the group of words no other words)?

Question

I have done some regex for 3 hours trying to figure this out.

I am trying to create an regular expression where I can find a group of specific words or numbers in any order, but this group can contain no other words or numbers unless it the selected words or number. Punctuation and spaces is fine

example:

the word I am searching for is one,three,four

one,three,four apple is red. should be match

four, three, one orange is orange. should be match

one,four,three the sky is blue. should be match

one,two,four,three should not be match because two is between the group of words

four,one,eight,green,three also not because eight and green is between the group of selected words

I have done some research and found that there is a way to match if a statement have a group of words in any order

Regex: I want this AND that AND that... in any order

The only problem with that is if I put a word that is not part of selected group in between the group that will also be consider match. I don't want that.

So I did some more research and found out how to selected specific words and found this

https://superuser.com/questions/903168/how-should-i-write-a-regex-to-match-a-specific-word

It works when one of the selected words are allow like the following below,

regex = (?:^|\W)one(?:$|\W)

I put "one" and it found one

I put "three two four one", It also find one

but when I put both expression together. I get this

^(?:^|\W)one(?:$|\W)(?:^|\W)three(?:$|\W)(?:^|\W)four(?:$|\W).*$

But it find nothing, what I do wrong?

three two four, It find nothing

The website I use is http://regexr.com/

Why `one,four,three the sky is blue.` isn't a valid sentence? — Casimir et Hippolyte, Apr 16 '17 at 14:49
It not valid. It a silly example. The main person is to find the one,four,three — Richard Twitty, Apr 16 '17 at 14:58
@CasimiretHippolyte it should be match, that was my mistake with edit, sorry — Shakiba Moshiri, Apr 16 '17 at 14:59
Add a tag for the language or tool in which you will write your regex. — Casimir et Hippolyte, Apr 16 '17 at 15:07

score 0 · Answer 1 · edited Jun 20 '20 at 09:12

0

based on what you comment me Yes the delimiter is always a comma, you can use negative look-ahead assertion to prevent more than 3 , if exist like:

`^\w+\s(.)\s\w+\s\1\s\w+(?:(?!\1).)*$`

here I picked up the delimiter by using capture-group by use can use it directly like:

`^\w+\s,\s\w+\s,\s\w+(?:(?!,).)*$`

or:

with string literal:

const regex = /^(?:one|three|four)\s*,\s*(?:one|three|four)\s*,\s*(?:one|three|four).*$/gm;
const str = `one,three,four apple is red.       match

four, three, one orange is orange. match

one,four,three the sky is blue.    match

one,two,four,three not match

four,one,eight,green,three no match`;
let m;

while ((m = regex.exec(str)) !== null) {
    // This is necessary to avoid infinite loops with zero-width matches
    if (m.index === regex.lastIndex) {
        regex.lastIndex++;
    }
    
    // The result can be accessed through the `m`-variable.
    m.forEach((match, groupIndex) => {
        console.log(`Found match, group ${groupIndex}: ${match}`);
    });
}

edited Jun 20 '20 at 09:12

Community

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answered Apr 16 '17 at 14:34

Shakiba Moshiri

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it not working for some reason. I try the first example, my test string is one,two,three. No match found – Richard Twitty Apr 16 '17 at 14:56
@LearnerMan Please be more specific. For what reason? – Shakiba Moshiri Apr 16 '17 at 14:57
^\w+\s*(.)\s*\w+\s*\1\s*\w+(?:(?!\1).)*$ was my regex and "one,two,three" is the string. I test it, but there no result – Richard Twitty Apr 16 '17 at 15:06
@LearnerMan Let me check it out – Shakiba Moshiri Apr 16 '17 at 15:11
@LearnerMan okey try thes: `(?:one|three|four)\s*,\s*(?:one|three|four)\s*,\s*(?:one|three|four).*` you need to use `|` operator – Shakiba Moshiri Apr 16 '17 at 15:11
@LearnerMan on regexr.com **turn on** g and m flag – Shakiba Moshiri Apr 16 '17 at 15:14
that would match "one,one,one" – Scott Weaver Apr 16 '17 at 15:18
yes, if the selected string "one,one,one" As long it match only the selected string I fine. – Richard Twitty Apr 16 '17 at 15:19
@LearnerMan okay if the answer is correct let me know so that I can fixed it that what pattern work for you: **first one** or **second one** – Shakiba Moshiri Apr 16 '17 at 15:22

score 0 · Answer 2 · answered Apr 16 '17 at 15:15

A combination of alternation | and negative lookaheads (?!..) with backreferences \n should do it:

You just match one legal value, then in the next sequence you do it again, but this time there's a negative lookahead to preclude what you already matched (which you represent with backreferences).

https://regex101.com/r/FwzBY9/1/

How to find group of words that is in random order(only the group of words no other words)?

2 Answers2

^\w+\s*(.)\s*\w+\s*\1\s*\w+(?:(?!\1).)*$

^\w+\s*,\s*\w+\s*,\s*\w+(?:(?!,).)*$

`^\w+\s(.)\s\w+\s\1\s\w+(?:(?!\1).)*$`

`^\w+\s,\s\w+\s,\s\w+(?:(?!,).)*$`