How to Check All Array Element Inside endswith()

Question

let array1 = ["?", "!", "."];
let array2 = ["live.", "ali!", "harp", "sharp%", "armstrong","yep?"];

console.log(array2.filter((x) => x.endsWith("?")));

The output is just: ['yep?']

Because the function endsWith() only checked for "?" as you see in the code. How do I loop the elements on the array1 (suffixes) inside the endsWith function so the output is:

['live.', 'ali!', 'yep?']

Answer 1

You could use a regex, and then match each element in the filter iteration against it.

/[?!.]$/ says match one of these things in the group ([?!.]) before the string ends ($).

const arr = ['live.', 'ali!', 'harp', 'sharp%', 'armstrong', 'yep?'];
const re = /[?!.]$/;

const out = arr.filter(el => el.match(re));
console.log(out);

Regarding your comment you can pass in a joined array to the RegExp constructor using a template string.

const query = ['.', '?', '!'];
const re = new RegExp(`[${query.join('')}]$`);

const arr = ['live.', 'ali!', 'harp', 'sharp%', 'armstrong', 'yep?'];

const out = arr.filter(el => el.match(re));
console.log(out);

Answer 2

You can use an inner .some() call to loop over your array1 and return true from that when you find a match instead of hardcoding the ?:

const array1 = ["?", "!", "."];
const array2 = ["live.", "ali!", "harp", "sharp%", "armstrong","yep?"];

const res = array2.filter((x) => array1.some(punc => x.endsWith(punc)));
console.log(res);

Above, when you return true (or a truthy value) from the .some() callback, the .some() method will return true, otherwise it will return false if you never return true, thus discarding it.