香草JS

ES2016：

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

ES6：

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

怎么运行的

some(..)根据测试函数检查数组的每个元素，如果数组的任何元素通过测试函数，则返回 true，否则返回 false。如果给定的参数存在于数组中indexOf(..) >= 0，则includes(..)两者都返回 true。

香草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;
    });
};

正如@loganfsmyth 所指出的，您可以在 ES2016 中将其缩短为

/**
 * @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));
};

或简单地作为 arr.some(v => haystack.includes(v));

如果要确定数组是否包含另一个数组中的所有项目，请替换some()为every() 或作为arr.every(v => haystack.includes(v));

ES6解决方案：

let arr1 = [1, 2, 3];
let arr2 = [2, 3];

let isFounded = arr1.some( ai => arr2.includes(ai) );

与它不同的是：必须包含所有值。

let allFounded = arr2.every( ai => arr1.includes(ai) );

希望，会有所帮助。

如果你不反对使用 libray，http ://underscorejs.org/有一个交集方法，它可以简化这个：

var _ = require('underscore');

var target = [ 'apple', 'orange', 'banana'];
var fruit2 = [ 'apple', 'orange', 'mango'];
var fruit3 = [ 'mango', 'lemon', 'pineapple'];
var fruit4 = [ 'orange', 'lemon', 'grapes'];

console.log(_.intersection(target, fruit2)); //returns [apple, orange]
console.log(_.intersection(target, fruit3)); //returns []
console.log(_.intersection(target, fruit4)); //returns [orange]

交叉函数将返回一个新数组，其中包含匹配的项目，如果不匹配，则返回空数组。

ES6（最快）

const a = ['a', 'b', 'c'];
const b = ['c', 'a', 'd'];
a.some(v=> b.indexOf(v) !== -1)

ES2016

const a = ['a', 'b', 'c'];
const b = ['c', 'a', 'd'];
a.some(v => b.includes(v));

下划线

const a = ['a', 'b', 'c'];
const b = ['c', 'a', 'd'];
_.intersection(a, b)

演示：https : //jsfiddle.net/r257wuv5/

jsPerf：https ://jsperf.com/array-contains-any-element-of-another-array