var target = ["apple","banana","orange"];
var checkArray = ["apple","banana","pineapple"];

var containsOneCommonItem = target.some(x => checkArray.some(y => y === x));`

["apple","grape"] //returns true;

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

["grape", "pineapple"] //returns false;

使用一些/findIndex和indexOf的组合怎么样？

所以类似这样：

var array1 = ["apple","banana","orange"];
var array2 = ["grape", "pineapple"];

var found = array1.some(function(v) { return array2.indexOf(v) != -1; });

为了使其更可读，可以将此功能添加到Array对象本身。

Array.prototype.indexOfAny = function (array) {
    return this.findIndex(function(v) { return array.indexOf(v) != -1; });
}

Array.prototype.containsAny = function (array) {
    return this.indexOfAny(array) != -1;
}

注意：如果您想对谓词执行某些操作，可以用另一个findIndex和谓词替换内部indexOf

您正在寻找两个数组之间的交集。你有两种主要的交叉点类型：“每个”和“一些”。让我举几个好例子：

每

let brands1 = ['Ford', 'Kia', 'VW', 'Audi'];
let brands2 = ['Audi', 'Kia'];
// Find 'every' brand intersection. 
// Meaning all elements inside 'brands2' must be present in 'brands1':
let intersectionEvery = brands2.every( brand => brands1.includes(brand) );
if (intersectionEvery) {
    const differenceList = brands1.filter(brand => !brands2.includes(brand));
    console.log('difference list:', differenceList);
    const commonList = brands1.filter(brand => brands2.includes(brand));
    console.log('common list:', commonList);
}

如果条件不满足（比如你在品牌中加入了“梅赛德斯”），那么“intersectionEvery”就不满足了——这将是错误的。

如果满足条件，它将把[“福特”、“大众”]列为区别，把[“起亚”、“奥迪”]列为了常见列表。

沙盒：https://jsfiddle.net/bqmg14t6/

SOME

let brands1 = ['Ford', 'Kia', 'VW', 'Audi'];
let brands2 = ['Audi', 'Kia', 'Mercedes', 'Land Rover'];
// Find 'some' brand intersection. 
// Meaning some elements inside 'brands2' must be also present in 'brands1':
let intersectionSome = brands2.some( brand => brands1.includes(brand) );
if (intersectionSome) {
    const differenceList = brands1.filter(brand => !brands2.includes(brand));
    console.log('difference list:', differenceList);
    const commonList = brands1.filter(brand => brands2.includes(brand));
    console.log('common list:', commonList);
}

我们在这里寻找一些常见的品牌，但不一定全部。

它将把[“福特”、“大众”]列为不同品牌，把[“起亚”、“奥迪”]列为了共同品牌。

沙盒：https://jsfiddle.net/zkq9j3Lh/

就我个人而言，我会使用以下功能：

var arrayContains = function(array, toMatch) {
    var arrayAsString = array.toString();
    return (arrayAsString.indexOf(','+toMatch+',') >-1);
}

“toString（）”方法将始终使用逗号分隔值。仅适用于基本类型。

这是一个有趣的案例，我认为我应该分享。

假设您有一个对象数组和一个选定过滤器数组。

let arr = [
  { id: 'x', tags: ['foo'] },
  { id: 'y', tags: ['foo', 'bar'] },
  { id: 'z', tags: ['baz'] }
];

const filters = ['foo'];

要将所选过滤器应用于此结构，我们可以

if (filters.length > 0)
  arr = arr.filter(obj =>
    obj.tags.some(tag => filters.includes(tag))
  );

// [
//   { id: 'x', tags: ['foo'] },
//   { id: 'y', tags: ['foo', 'bar'] }
// ]

ES6溶液：

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

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

与之不同：必须包含所有值。

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

希望，会有所帮助。