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Cómo escapar adecuadamente de los caracteres en esta construcción de expresiones regulares

Estoy construyendo una función rstrip ingenua. Y funciona correctamente así cuando codifico los caracteres reales:

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

Sin embargo, si trato de usar la variable chars , tengo problemas para escapar:

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

¿Cuál sería la forma correcta de 'escapar' de la variable para que el valor !r reemplace la variable chars en el método de replace ?

about 4 years ago · Juan Pablo Isaza
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 const rstrip = (s, chars) => s.replace(new RegExp(`[${chars}]+$`, 'g'), ""); console.log(rstrip("Hello!!!!r", '!r'))

about 4 years ago · Juan Pablo Isaza Report
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