JS Regex: (a|b) Matches the a or the b part of the subexpression. Different behaviour when a and b switched

Question

I am trying to replace 2% with ''(empty) using regexp concept. If input string is either % or 2%, it should be replaced with ''(empty):

const str = "2%";

console.log(`2%`.replace(/^\d%$|\d(?=%)/, ''));
console.log(`2%`.replace(/\d(?=%)|^\d%$/, ''));

(a|b) Matches the a or the b part of the subexpression.

"2%".replace(/^\d%$|\d(?=%)/, ''). This works well.

But, "2%".replace(/\d(?=%)|^\d%$/, '') does not.

So what is your question and problem? You appear to have working source code.... — NewToJS, Mar 09 '19 at 20:34
The left side of the `|` matches, so that's what's replaced. What exactly is it that's surprising here? — Pointy, Mar 09 '19 at 20:36

score 2 · Accepted Answer · answered Mar 09 '19 at 20:39

2

The difference is in what the regex tries to match first. The left expression takes precedence. The right is only tried when the left one fails to match.

answered Mar 09 '19 at 20:39

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JS Regex: (a|b) Matches the a or the b part of the subexpression. Different behaviour when a and b switched

1 Answers1