do loose equality comparision using array.includes

Question

Array.includes does a strict comparison of array elements.

var array1 = [1, 2, 3];

console.log(array1.includes(2)); // return true

console.log(array1.includes("2")); // return false

But I want the result to return true in second case as well.My main goal is to know whether an array contains an element. Please suggest how this can be achieved.

Answer 1

You can't use includes, since it always uses strict equality; you can use some instead:

console.log(array1.some(e => e == "2"));

some calls the callback you provide for the elements, in order, until your callback returns a truthy value, in which case some stops looping and returns true. If your callback never returns a truthy value (including because it was never called because the array was empty), some returns false.

some was added in ES5 (2009).

Live example:

const array1 = [1, 2, 3];
console.log(array1.some(e => e == "2"));

Answer 2

If the array is always formed by numbers and you are sure to pass a string to be checked, you could also use parseInt():

console.log(array1.includes(parseInt("2")));

Please note that this would work also if you were to pass an integer to the parseInt but that is is not correct since parseInt expect a string as a parameter.