Regex remove duplicate adjacent characters in javascript

Question

I've been struggling getting my regex function to work as intended. My goal is to iterate endlessly over a string (until no match is found) and remove all duplicate, adjacent characters. Aside from checking if 2 characters (adjacent of each other) are equal, the regex should only remove the match when one of the pair is uppercase.

e.g. the regex should only remove 'Xx' or 'xX'.

My current regex only removes matches where a lowercase character is followed by any uppercase character.

(.)(([a-z]{0})+[A-Z])

How can I implement looking for the same adjacent character and the pattern of looking for an uppercase character followed by an equal lowercase character?

is there a need for regex here?

Kobe
– Kobe

2019-07-23 07:29:53 +00:00
Commented Jul 23, 2019 at 7:29 — Kobe
– Kobe, Commented Jul 23, 2019 at 7:29

CertainPerformance · Accepted Answer · 2019-07-23 07:36:08Z

3

You'd either have to list out all possible combinations, eg

aA|Aa|bB|Bb...

Or implement it more programatically, without regex:

let str = 'fooaBbAfoo';
outer:
while (true) {
  for (let i = 0; i < str.length - 1; i++) {
    const thisChar = str[i];
    const nextChar = str[i + 1];
    if (/[a-z]/i.test(thisChar) && thisChar.toUpperCase() === nextChar.toUpperCase() && thisChar !== nextChar) {
      str = str.slice(0, i) + str.slice(i + 2);
      continue outer;
    }
  }
  break;
}
console.log(str);

answered Jul 23, 2019 at 7:36

CertainPerformance

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Lyzard Kyng · Accepted Answer · 2019-07-23 10:51:16Z

3

Looking for the same adjacent character: /(.)\1/

Looking for an uppercase character followed by an equal lowercase character isn't possible in JavaScript since it doesn't support inline modifiers. If they were regex should be: /(.)(?!\1)(?i:\1)/, so it matches both 'xX' or 'Xx'

answered Jul 23, 2019 at 10:51

Lyzard Kyng

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