Javascript Question posted in
A very good W3school tutorial can be found here.

How to migrate this regex to JavaScript

I have this regex that works perfectly with PHP:

$que = preg_replace("/(?<=^| )([a-z])\1*(?= |$)K|([a-z])(?=\2)/","$3",$que);

This regex removes repeated chars inside strings (for example, axxd becomes axd however xxx will still become xxx). The problem that I am facing, is because it does not work with JS (I think the negative lookbehind does not work with JS).

More examples:

  1. the string aaa baaab xxx would become aaa bab xxx
  2. the string ahhj aaab cc iiik would become ahj ab cc ik

Do you have a solution for this that is at least a little efficient? I mean, I will probably use this regex on a string with 1k chars, so if the regex is not efficient, the browser may freeze.

Tags:

2

Answers


  1. 0

    The negative lookbehind is not likely to be your issue as they are supported on almost all current release browsers. However JavaScript regex doesn’t recognise K as a meta sequence but rather as a literal K. You can work around that using this regex:

    b([a-z])1+(?!1|b)|(?<=([a-z]))((?!2)[a-z])3+

    This matches either b([a-z])1+(?!1|b):

    • b : word boundary
    • ([a-z]) : a letter, captured in group 1
    • 1+ : one or more repetitions of the captured letter
    • (?!1|b) : lookahead assertion that the next location is not another repetition of the captured letter or a word boundary

    or (?<=([a-z]))((?!2)[a-z])3+:

    • (?<=([a-z])) : a positive lookbehind for a letter, captured in group 2
    • ((?!2)[a-z]) : another letter which is not the same as the previously captured letter, captured in group 3
    • 3+ : one of more repetitions of the captured letter

    The first part of the regex will capture repeated letters at the beginning of a word; the second part captures repeated letters in the middle or at the end of a word.

    You can then replace with $1$3 which will replace any repeated letter matched by the regex with just a single copy of itself.

    Regex demo on regex101

    In JavaScript:

    console.log('aaa baaab xxx fjjj'.replace(/b([a-z])1+(?!1|b)|(?<=([a-z]))((?!2)[a-z])3+/g, '$1$3'))
console.log('ahhj aaab cc iiik'.replace(/b([a-z])1+(?!1|b)|(?<=([a-z]))((?!2)[a-z])3+/g, '$1$3'))
console.log('bbb ahhj aaab cc iiik xxx fjjj baaaaaab yyyaaa'.replace(/b([a-z])1+(?!1|b)|(?<=([a-z]))((?!2)[a-z])3+/g, '$1$3'))

    PHP demo on 3v4l.org

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  2. 0 
    let result = str.replace(/b((.)2+)b|(.)3+/g, '$1$3');

    demo

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