Regex to find even occurrence of a character in a string every time it repeats in a string in JavaScript

Issue

Regex to find even occurrence of a character in a string every time it repeats in a string

Example:

YYMMDD-YYYY-DD true
YYMDD false

Y,M,D are case sensitive Y,M,D can appear multiple time at multiple postion in pairs in string but each pair must be even.

I have applied the above even check using for loop but have to replace for loop with regex I have also tried it with the below regex but it didn’t worked

if (result.match(/M{2,}/) || result.match(/D{2,}/) || result.match(/Y{2,}/)) {}

solution using for loop which i have implemented

Solution

[ 'YYMMDD-YYYY-DD', 'YYMMDD', 'YYMDD', 'Y', 'YY', 'YYY', 'YYYY', 'XX' ].forEach(str => {
  let ok = /^(?:([YMD])\1-?)+$/.test(str);
  console.log(str + ' => ' + ok);
});

Output:

YYMMDD-YYYY-DD => true
YYMMDD => true
YYMDD => false
Y => false
YY => true
YYY => false
YYYY => true
XX => false

Explanation of regex:

• ^ -- anchor at start of string
• (?: -- non-capture group start
• ([YMD]) -- capture group 1: single alpha character
• \1 -- repeat capture group 1
• -? -- optional - (if needed change to a character class to allow additional chars)
• )+ -- non-capture group end, repeat 1+ times
• $ -- anchor at end

Answered By - Peter Thoeny