Question

我有一个对象：

const animals = [
    {name: 'Fluffy', species: 'cat'},
    {name: 'Crinkle', species: 'rabbit'},
    {name: 'Wally', species: 'dog'},
    {name: 'Roo', species: 'dog'},
    {name: 'Felix', species: 'cat'},
]

我想使用诸如filter（）方法之类的高阶函数来获取动物对象的数组，并返回仅包含所有狗名的数组，即["Wally", "Roo"]。目前，我的代码返回一个数组，其中包含带有对象dog的整个对象。见下文：

const dogArray = animals.filter(function(animal) {
  return animal.species === 'dog';
 })

return dogArray;

// returns
// [{name: "Wally", species: "dog"}, 
// { name: "Roo", species: "dog"}]

Answer 1

只需将过滤后的数组的元素映射到其name属性：

const animals = [
  {name: 'Fluffy', species: 'cat'},
  {name: 'Crinkle', species: 'rabbit'},
  {name: 'Wally', species: 'dog'},
  {name: 'Roo', species: 'dog'},
  {name: 'Felix', species: 'cat'},
]

const dogArray = animals.filter(animal => animal.species === 'dog');

console.log(dogArray.map(dog => dog.name));

或将两者合并为一个约简：

const animals = [
  {name: 'Fluffy', species: 'cat'},
  {name: 'Crinkle', species: 'rabbit'},
  {name: 'Wally', species: 'dog'},
  {name: 'Roo', species: 'dog'},
  {name: 'Felix', species: 'cat'},
]

let dogArray = animals.reduce((dogs, animal) => {
  if (animal.species === "dog") dogs.push(animal.name);
  return dogs;
}, []);

console.log(dogArray)

Answer 2

您可以使用destructuring映射属性。

const
    animals = [{ name: 'Fluffy', species: 'cat' }, { name: 'Crinkle', species: 'rabbit' }, { name: 'Wally', species: 'dog' }, { name: 'Roo', species: 'dog' }, { name: 'Felix', species: 'cat' }]
    dogArray = animals
        .filter(({ species }) => species === 'dog')
        .map(({ name }) => name);

console.log(dogArray);

Answer 3

创建一个空数组，使用dogArray循环遍历现有for，将名称推入新数组，然后返回新数组。

const dogArray = animals.filter(function(animal) {
  return animal.species === 'dog';
 })

let dogNames = [];

for (let i in dogArray) {
  dogNames.push(dogArray[i].name);
}

return dogNames;

Answer 4

/application-a/

Answer 5

您可以使用.filter()，然后使用.map()：

const animals = [
  {name: 'Fluffy', species: 'cat'},
  {name: 'Crinkle', species: 'rabbit'},
  {name: 'Wally', species: 'dog'},
  {name: 'Roo', species: 'dog'},
  {name: 'Felix', species: 'cat'},
];

const dogNames = animals
  .filter(animal => animal.species === 'dog')
  .map(dog => dog.name);

console.log(dogNames);

使用更高阶的函数，如果另一个值为true，则返回一个对象值（JavaScript）

5 个答案: