Matching the last best word using regex js

Question

I have a regex for matching the text within the brackets. For example the regex https://regex101.com/r/TvweUj/3

/\b(\w)[-'\w]* (?:[-",/\\*&'\w]* ){1,}\(\1[A-Z]{1,}\)/gi

matches MIDI The USB Device Class Definition for MIDI Devices transmits Music Instrument Digital Interface (MIDI). instead of only matching the last 4 words Music Instrument Digital Interface

How do I change my regex to match the recent matching instead from the MIDI The USB Dev*****

Answer 1

You might use 4 capturing groups with a positive lookahead asserting 4 backreferences to match the uppercase chars between the parenthesis:

\b([A-Z])\w+ ([A-Z])\w+ ([A-Z])\w+ ([A-Z])\w+(?= \(\1\2\3\4\))

Regex demo

Instead of using \w only, you could use the character classes that you use in the question like [-",/\\*&'\w]*

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A more broad pattern could be repeating an uppercase char followed by 1+ word chars \w+ (or use \w* to repeat 0+ word chars) and assert that what follows is only uppercase chars between parenthesis.

\b[A-Z]\w+(?: [A-Z]\w+)*(?= \([A-Z]+\))

Regex demo

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If the number of chars are variable that you want to match between the parenthesis and they should match with the number of words before, you could use 2 capturing groups and compare the amount of splitted words with the number of uppercase chars between the parenthesis.

let pattern = /\b([A-Z][a-z]*(?: [A-Z][a-z]*)*) \(([A-Z]+)\)/;
let compare = (ar1, ar2) =>
  ar1.length === ar2.length && ar1.every(
    (value, index) => value === ar2[index].charAt(0)
  );
[
  "transmits Music Instrument Digital Interface (MIDI).",
  "transmits Music Instrument Digital Interface (MADI).",
  "transmits Music Instrument Digital Interface (MID)."
].forEach(s => {
  let m = s.match(pattern);
  let res = compare(m[2].split(''), m[1].split(' ')) ? "Ok -> " : "Not ok -> ";
  console.log(res + s);
})