How to properly escape the characters in this regex construction

Question

I am building a naive rstrip function. And it works properly like this when I hardcode the actual characters:

const rstrip = (s, chars) => s.replace(/[r!]+$/g, "");
console.log(rstrip("Hello!!!!r", '!r'))

However, if I try and use the chars variable, I have trouble escaping it:

const rstrip = (s, chars) => s.replace(/[chars]+$/g, "");
console.log(rstrip("Hello!!!!r", '!r'))

What would be the proper way to 'escape' the variable so that the value !r replaces the chars variable in the replace method?

Answer 1

const rstrip = (s, chars) => s.replace(new RegExp(`[${chars}]+$`, 'g'), "");
console.log(rstrip("Hello!!!!r", '!r'))