我正在尝试遍历一个数组并找到在属性中具有所述关键字的每个对象

Question

So I want to set this variable to all of the objects in an array in which its category has a certain keyword. This is what I have so far

  let tempMensProducts = tempClothingProducts
          .filter(obj => obj.category.includes('Mens'))

export const clothingProducts = [
  {
    id: 1,
    title: 'COMME DES GARCONS TEE',
    img: 'img/product-1.png',
    img2: 'img/product-1-1.png',
    img3: 'img/product-1-2.png',
    img4: 'img/product-1-3.png',
    luxury: 'All Luxury items are inspected to verify authenticity',
    price: 200,
    info: ' COMME DES GARCONS PLAY BASIC LOGO TEE',
    inCart: false,
    count: 0,
    total: 0,
    fabric: '100% Cotton',
    category: 'Mens Fashion'
  }
]

Answer 1

You need to pass a callback to map to use it - but here, map isn't required at all. Just use filter .

let tempMensProducts = tempClothingProducts.filter(({ category }) => category.includes("Mens"));

If you try to use map without a callback, it'll result in an error as it tries to call its argument, and if you don't pass one, it'll try to do undefined() .

 [1, 2, 3].map();

Answer 2

You can split the array by whitespaces and then apply array contains there to sure that it will give you correct result

I mean to say it should not return "Women" when keyword is "men"

 let clothingProducts = [ { id: 1, title: 'COMME DES GARCONS TEE', img: 'img/product-1.png', img2: 'img/product-1-1.png', img3: 'img/product-1-2.png', img4: 'img/product-1-3.png', luxury: 'All Luxury items are inspected to verify authenticity', price: 200, info: ' COMME DES GARCONS PLAY BASIC LOGO TEE', inCart: false, count: 0, total: 0, fabric: '100% Cotton', category: 'Mens Fashion' }, { id: 2, title: 'COMME DES GARCONS TEE', img: 'img/product-1.png', img2: 'img/product-1-1.png', img3: 'img/product-1-2.png', img4: 'img/product-1-3.png', luxury: 'All Luxury items are inspected to verify authenticity', price: 200, info: ' COMME DES GARCONS PLAY BASIC LOGO TEE', inCart: false, count: 0, total: 0, fabric: '100% Cotton', category: 'woMens Fashion' } ] let filter = clothingProducts.filter(c => c.category.split(" ").includes("Mens")); console.log(filter);