基于2个值过滤嵌套对象

Question

因此，如果链接包含/罗马/它将列出所有包含/罗马/我设法做到的页面，我想创建一个基于链接返回数据的筛选器。但是我也试图创建一个基于城市市场的过滤器，例如，如果链接包含/罗马/超市/我想过滤掉页面，只返回包含城市和市场的页面，在我的最后一个链接示例中，这意味着id 1和2。

首先，我认为我需要遍历所有的页面，在我的例子中是3，然后再循环一次，让每个城市检查它是否包含那个城市和超市，并根据它返回过滤后的数据。我试过几样东西，但我在这里挠头。

 pages: [
    {
      id: 1,
      name: 'Meals',
      startDate: '2022-11-09 10:32:00',
      endDate: '2022-11-16 10:32:00',
      cities: {
        rome: ['supermarket'],
      },
    },
    {
      id: 2,
      name: 'Deals',
      startDate: '2022-11-24 11:01:00',
      endDate: '2022-12-01 11:01:00',
      cities: {
        napoli: ['supermarket', 'minimarket'],
        rome: ['supermarket'],
      },
    },
    {
      id: 3,
      name: 'Toys',
      startDate: '2022-11-24 11:01:00',
      endDate: '2022-12-01 11:01:00',
      cities: {
        rome: ['minimarket'],
        venice: ['supermarket', 'minimarket'],
      },
    }
]

城市->的过滤器

      const pagesFilterCity = pages.filter((item) => {
        return item.cities.hasOwnProperty(city); // where city is 'rome'
      });

我试过的->

      const pagesFilterMarket = pages.filter((item) => {
        return Object.values(item.cities)
          .flat()
          .some((item) => item === market); // where market is 'supermarket'
      });

但是这只会从每个城市循环第一个实例，这意味着3次。

想要的结果:在链接中有/罗马/超市之后，我想创建一个过滤器，它应该从上面的-> 页面中返回以下数据--注意，从我已经做过的链接中提取数据，这不是我的问题

 pages: [
    {
      id: 1,
      name: 'Meals',
      startDate: '2022-11-09 10:32:00',
      endDate: '2022-11-16 10:32:00',
      cities: {
        rome: ['supermarket'],
      },
    },
    {
      id: 2,
      name: 'Deals',
      startDate: '2022-11-24 11:01:00',
      endDate: '2022-12-01 11:01:00',
      cities: {
        napoli: ['supermarket', 'minimarket'],
        rome: ['supermarket'],
      },
    }
] // wanted result

Answer 1

据我所知，你正在检查两件事：

1. 城市对象是否包含所需的城市？
2. 这个城市有特定的市场吗？

这可以通过类似于以下几点来实现：

​

const pages = [{
    id: 1,
    name: 'Meals',
    startDate: '2022-11-09 10:32:00',
    endDate: '2022-11-16 10:32:00',
    cities: {
        rome: ['supermarket'],
    },
},
{
    id: 2,
    name: 'Deals',
    startDate: '2022-11-24 11:01:00',
    endDate: '2022-12-01 11:01:00',
    cities: {
        napoli: ['supermarket', 'minimarket'],
        rome: ['supermarket'],
    },
},
{
    id: 3,
    name: 'Toys',
    startDate: '2022-11-24 11:01:00',
    endDate: '2022-12-01 11:01:00',
    cities: {
        rome: ['minimarket'],
        venice: ['supermarket', 'minimarket'],
    },
},
];

const search = (location, market) => {
    return pages.filter((record) => 
        // Check if the cities property has a value
        record.cities[location] && 
        // Check if the specific city has the given market
        record.cities[location].some((type) => type === market)
    );
};

console.log(search('rome', 'supermarket'));

​

如果你想做一个额外的检查，并在安全方面，你也可以包括

return pages.filter((record) => 
    record.cities[location] && 
    record.cities[location].length &&
    record.cities[location].some((type) => type === market)
);

Answer 2

您可以使用/要拥有城市和服务将字符串拆分为两部分。然后，您只需检查服务是否为空或城市数组是否包含所请求的服务。

​

const pages = [{
    id: 1,
    name: 'Meals',
    startDate: '2022-11-09 10:32:00',
    endDate: '2022-11-16 10:32:00',
    cities: {
      rome: ['supermarket'],
    },
  },
  {
    id: 2,
    name: 'Deals',
    startDate: '2022-11-24 11:01:00',
    endDate: '2022-12-01 11:01:00',
    cities: {
      napoli: ['supermarket', 'minimarket'],
      rome: ['supermarket'],
    },
  },
  {
    id: 3,
    name: 'Toys',
    startDate: '2022-11-24 11:01:00',
    endDate: '2022-12-01 11:01:00',
    cities: {
      rome: ['minimarket'],
      venice: ['supermarket', 'minimarket'],
    },
  }
]

function filter(search) {
  const [city, service] = search.split('/');
  return pagesFilterCity = pages.filter((item) => {
    return item.cities.hasOwnProperty(city) && (service === '' || item.cities[city].includes(service)); // where city is rome for example
  });
}
console.log(filter('rome/supermarket'))

​

Answer 3

您可以将Array.prototype.filter()与可选链(？)和Array.prototype.includes()结合使用。

代码：

​

const pages = [{id: 1,name: 'Meals',startDate: '2022-11-09 10:32:00',endDate: '2022-11-16 10:32:00',cities: {rome: ['supermarket'],},},{id: 2,name: 'Deals',startDate: '2022-11-24 11:01:00',endDate: '2022-12-01 11:01:00',cities: {napoli: ['supermarket', 'minimarket'],rome: ['supermarket'],},},{id: 3,name: 'Toys',startDate: '2022-11-24 11:01:00',endDate: '2022-12-01 11:01:00',cities: {rome: ['minimarket'],venice: ['supermarket', 'minimarket']}}]

const result = pages.filter(o => o.cities['rome']?.includes('supermarket'))

console.log(result)

​