Question

我有一个数组如：

const cars = [
  { id: 23423, brand: 'bmw', doors: 2, color: 'red' },
  { id: 23452, brand: 'volvo', doors: 4, color: 'gray' },
  { id: 97456, brand: 'citroen', doors: 4, color: 'black' },
  { id: 45784, brand: 'dodge', doors: 2, color: 'red' },
  { id: 23452, brand: 'ferrari', doors: 2, color: 'red' },
  { id: 23522, brand: 'bmw', doors: 2, color: 'blue' }
];

这里我有一个id列表，例如：[45784, 23522]

我想找到使用map，filter或reduce的最佳方法，使用ids数组检索第一个数组中的项目。

由于

Answer 1

简单如下：

const 
  cars = [
  { id: 23423, brand: 'bmw', doors: 2, color: 'red' },
  { id: 23452, brand: 'volvo', doors: 4, color: 'gray' },
  { id: 97456, brand: 'citroen', doors: 4, color: 'black' },
  { id: 45784, brand: 'dodge', doors: 2, color: 'red' },
  { id: 23452, brand: 'ferrari', doors: 2, color: 'red' },
  { id: 23522, brand: 'bmw', doors: 2, color: 'blue' }
],
  ids = [45784, 23522];

let arr = cars.filter(elem => ids.includes(elem.id));

console.log(arr);

Answer 2

您可以使用Array#includes进行过滤。

此提案使用destructuring assignment作为对象的id属性。

＆＃13;

var cars = [{ id: 23423, brand: 'bmw', doors: 2, color: 'red' }, { id: 23452, brand: 'volvo', doors: 4, color: 'gray' }, { id: 97456, brand: 'citroen', doors: 4, color: 'black' }, { id: 45784, brand: 'dodge', doors: 2, color: 'red' }, { id: 23452, brand: 'ferrari', doors: 2, color: 'red' }, { id: 23522, brand: 'bmw', doors: 2, color: 'blue' }],
    ids = [45784, 23522],
    result = cars.filter(({ id }) => ids.includes(id));

console.log(result);

＆＃13;

Answer 3

从ids数组转换Set，并使用Array.filter()仅包含集合中存在的项目：

＆＃13;

const cars = [
  { id: 23423, brand: 'bmw', doors: 2, color: 'red' },
  { id: 23452, brand: 'volvo', doors: 4, color: 'gray' },
  { id: 97456, brand: 'citroen', doors: 4, color: 'black' },
  { id: 45784, brand: 'dodge', doors: 2, color: 'red' },
  { id: 23452, brand: 'ferrari', doors: 2, color: 'red' },
  { id: 23522, brand: 'bmw', doors: 2, color: 'blue' }
];

const ids = [45784, 23522];

const idsSet = new Set(ids);
const result = cars.filter(({ id }) => idsSet.has(id));

console.log(result);

＆＃13;

通过ID从列表中检索多个项目

3 个答案: