Excluding Punctuation Marks in JavaScript include method

Question

Is there a better way of checking if a word exists in an array of strings without including Punctuation Marks?

what I have tried,

const sArray=['Lorem','Ipsum','typesetting-','industry.','Ipsum?','has' ]


console.log(sArray.toString().replaceAll(".","").includes("industry")) //true

console.log(sArray.includes("industry")) //false

`sArray.some(e => e.includes("industry"))`. Pay attention to the fact that the `includes` here is `String.prototype.includes`, not `Array.prototype.includes`. — Gerardo Furtado, Oct 10 '22 at 10:25
As @GerardoFurtado said or, in case you need to manipulate the string when it is present, you can use `Array.prototype.find()` (`const foundItem = sArray.find(e => e.includes("industry")); if (foundItem !== -1) { /* manipulate the string */ }`) — secan, Oct 10 '22 at 10:40

score 0 · Answer 1 · answered Oct 10 '22 at 10:31

0

Something like this?

sArray.some(str => str.includes('industry'))

answered Oct 10 '22 at 10:31

Morteza Gholami

59
3

score 0 · Answer 2 · answered Oct 10 '22 at 10:32

If you do a lot of operations on the array, I suggest creating a new reference of the array after replacing, then dealing with the new array.

const sArray=['Lorem','Ipsum','typesetting-','industry.','Ipsum?','has' ]

const trasformedArray = sArray.map(i => i.replaceAll('.', ''));

console.log(trasformedArray.includes("industry"))

Excluding Punctuation Marks in JavaScript include method

2 Answers2