Check if an array contains any element of another array in JavaScript

Question

I have a target array ["apple","banana","orange"], and I want to check if other arrays contain any one of the target array elements.

For example:

["apple","grape"] //returns true;  ["apple","banana","pineapple"] //returns true;  ["grape", "pineapple"] //returns false; 

How can I do it in JavaScript?

Answer 1

Vanilla JS

ES2016:

const found = arr1.some(r=> arr2.includes(r)) 

ES6:

const found = arr1.some(r=> arr2.indexOf(r) >= 0) 

How it works

some(..) checks each element of the array against a test function and returns true if any element of the array passes the test function, otherwise, it returns false. indexOf(..) >= 0 and includes(..) both return true if the given argument is present in the array.

Answer 2

vanilla js

/**  * @description determine if an array contains one or more items from another array.  * @param {array} haystack the array to search.  * @param {array} arr the array providing items to check for in the haystack.  * @return {boolean} true|false if haystack contains at least one item from arr.  */ var findOne = function (haystack, arr) {     return arr.some(function (v) {         return haystack.indexOf(v) >= 0;     }); }; 

As noted by @loganfsmyth you can shorten it in ES2016 to

/**  * @description determine if an array contains one or more items from another array.  * @param {array} haystack the array to search.  * @param {array} arr the array providing items to check for in the haystack.  * @return {boolean} true|false if haystack contains at least one item from arr.  */ const findOne = (haystack, arr) => {     return arr.some(v => haystack.includes(v)); }; 

or simply as arr.some(v => haystack.includes(v));

If you want to determine if the array has all the items from the other array, replace some() to every() or as arr.every(v => haystack.includes(v));