Check if array contains all elements of another array

Question

I want a function that returns true if and only if a given array includes all the elements of a given "target" array. As follows.

const target = [ 1, 2, 3,    ]; const array1 = [ 1, 2, 3,    ]; // true const array2 = [ 1, 2, 3, 4, ]; // true const array3 = [ 1, 2,       ]; // false 

How can I accomplish the above result?

Answer 1

You can combine the .every() and .includes() methods:

let array1 = [1,2,3],     array2 = [1,2,3,4],     array3 = [1,2];  let checker = (arr, target) => target.every(v => arr.includes(v));  console.log(checker(array2, array1));  // true console.log(checker(array3, array1));  // false

Answer 2

The every() method tests whether all elements in the array pass the test implemented by the provided function. It returns a Boolean value. Stands to reason that if you call every() on the original array and supply to it a function that checks if every element in the original array is contained in another array, you will get your answer. As such:

const ar1 = ['a', 'b']; const ar2 = ['c', 'd', 'a', 'z', 'g', 'b'];  if(ar1.every(r => ar2.includes(r))){   console.log('Found all of', ar1, 'in', ar2); }else{   console.log('Did not find all of', ar1, 'in', ar2); }