Regex lookbehind workaround for Javascript?

Question

I am terrible at regex so I will communicate my question a bit unconventionally in the name of trying to better describe my problem.

var TheBadPattern = /(\d{2}:\d{2}:\d{2},\d{3})/;
var TheGoodPattern = /([a-zA-Z0-9\-,.;:'"])(?:\r\n?|\n)([a-zA-Z0-9\-])/gi;

// My goal is to then do this
inputString = inputString.replace(TheGoodPattern, '$1 $2);

Question: I want to match all the good patterns and do the subsequent find/replace UNLESS they are proceeded by the bad pattern, any ideas on how? I was able to accomplish this in other languages that support lookbehind but I am at a loss without it? (ps: from what I understand, JS does not support lookahead/lookbehind or if you prefer, '?>!', '?<=')

Answer 1

JavaScript does support lookaheads. And since you only need a lookbehind (and not a lookahead, too), there is a workaround (which doesn't really aid the readability of your code, but it works!). So what you can do is reverse both the string and the pattern.

inputString = inputString.split("").reverse().join("");
var pattern = /([a-z0-9\-])(?:\n\r?|\r)([a-z0-9\-,.;:'"])(?!\d{3},\d{2}:\d{2}:\d{2})/gi
inputString = inputString.replace(TheGoodPattern, '$1 $2');
inputString = inputString.split("").reverse().join("");

Note that you had redundantly used the upper case letters (they are being taken care of the i modifier).

I would actually test it for you if you supplied some example input.