尝试将数组与嵌套在对象中的另一个数组进行比较答案

【问题标题】：Trying to compare an array with another array nested in an object尝试将数组与嵌套在对象中的另一个数组进行比较
【发布时间】：2021-05-29 07:53:58
【问题描述】：

我在这个问题上已经卡住了一段时间。如果有人对我如何解决这个问题有任何指导。

此函数返回访问给定用户愿望清单中任何公园的所有用户名。

getUsersForUserWishlist(users, "karah.branch3"); //> ["dwayne.m55"]
getUsersForUserWishlist(users, "dwayne.m55"); //> []

const parks = [{
    id: 1,
    name: "Acadia",
    areaInSquareKm: 198.6,
    location: {
      state: "Maine"
    },
  },
  {
    id: 2,
    name: "Canyonlands",
    areaInSquareKm: 1366.2,
    location: {
      state: "Utah"
    },
  },
  {
    id: 3,
    name: "Crater Lake",
    areaInSquareKm: 741.5,
    location: {
      state: "Oregon"
    },
  },
  {
    id: 4,
    name: "Lake Clark",
    areaInSquareKm: 10602,
    location: {
      state: "Alaska"
    },
  },
  {
    id: 5,
    name: "Kenai Fjords",
    areaInSquareKm: 2710,
    location: {
      state: "Alaska"
    },
  },
  {
    id: 6,
    name: "Zion",
    areaInSquareKm: 595.9,
    location: {
      state: "Utah"
    },
  },
];

const users = {
  "karah.branch3": {
    visited: [1],
    wishlist: [4, 6],
  },
  "dwayne.m55": {
    visited: [2, 5, 1],
    wishlist: [],
  },
  thiagostrong1: {
    visited: [5],
    wishlist: [6, 3, 2],
  },
  "don.kim1990": {
    visited: [2, 6],
    wishlist: [1],
  },
};

function getUsersVisitedForUserWishlist(users, username) {}

【问题讨论】：

标签： javascript arrays function object oop

【解决方案1】：

这解决了你的问题

 function getUsersForUserWishlist(users1, userName) {
        //Retreive the user wishlist
        var wishlistInd = users1[arguments[1]]["wishlist"];
        var arrayRes = [];
        //loop throght users array
        for (const [key, value] of Object.entries(users1)) {
           var filteredArray = [];
          if (  `${key}` !== userName ) {
           var filteredArray = value['visited'].filter(value => 
            wishlistInd.includes(value));
            if  (filteredArray.length != 0 ) {
              arrayRes.push( `${key}`);
            }
          }
        }
        return arrayRes;
        }

【讨论】：

【解决方案2】：

如果我正确理解您的函数必须返回（如下例所示）而不是您的示例中提到的：

getUsersForUserWishlist(users, "karah.branch3"); //> ["don.kim1990"]
getUsersForUserWishlist(users, "dwayne.m55"); //> []

【讨论】：