如果对象具有来自不同数组对象键值的键，则仅返回该键

Question

我有2个数组。

一个数组包含一些人员对象，另一个数组包含具有名称键的对象，该名称键保存人员对象所需的值。

到目前为止，我的解决方案但没有得到任何运气......

在映射人员数组时，如何仅返回人员的某​​些属性？不是整个人员对象

const customContactValues = people.map((person) => { return valuesNeeded.filter((el) => (el.name in person) ? person[el.name] : ""  ) })

console.log(customContactValues)

这是我的人员数组

    const people = [
    {
        "foods": {
            "favoriteFood": "Ice cream"
        },
        "firstName": "John",
        "lastName": "Doe",
        "age": 30

    },
    {
        "foods": {
            "favoriteFood": "pizza"
        },
        "firstName": "Jane",
        "lastName": "Doe",
        "age": 39
    },
    {
        "foods": {
            "favoriteFood": "pbj"
        },
        "firstName": "Kevin",
        "lastName": "Baker",
        "age": 22

    },
    {
        "foods": {
            "favoriteFood": "pushpops"
        },
        "firstName": "ernie",
        "lastName": "duncan",
        "age": 29
    },
]

的键，如下所示

const valuesNeeded = [ 
  { name: 'firstName' },
  { name: 'lastName' },
  { name: 'favoriteFood' }
]

这是值数组，其中包含我需要从人员数组中获取和输出

const desiredResult = [
{firstName: "John", lastName: "Doe", favoriteFood: "Ice cream"},
{firstName: "Jane", lastName: "Doe", favoriteFood: "Pizza"},
{firstName: "Kevin", lastName: "Baker", favoriteFood: "pbj"},
{firstName: "ernie", lastName: "duncan", favoriteFood: "pushpops"}
]

Answer 1

filter 仅根据某些条件过滤数组中的元素，但在您的情况下，我们不想过滤元素，我们只想从现有数组创建新的对象数组，因此 map 功能是一个好的开始。

第二个问题是对象可以包含嵌套对象，嵌套对象可能包含所需的键值对，因此为了检索它们，如果我们没有直接在对象中找到键值，我们可以递归地查找对象的值。

由于我们不需要数组中每个对象的所有键值，因此我们可以创建一个新对象或从现有对象中删除，如果需要进一步处理，保留原始对象是一个很好且安全的选择。

const people = [{
    "foods": {
      "favoriteFood": "Ice cream"
    },
    "firstName": "John",
    "lastName": "Doe",
    "age": 30

  },
  {
    "foods": {
      "favoriteFood": "pizza"
    },
    "firstName": "Jane",
    "lastName": "Doe",
    "age": 39
  },
  {
    "foods": {
      "favoriteFood": "pbj"
    },
    "firstName": "Kevin",
    "lastName": "Baker",
    "age": 22

  },
  {
    "foods": {
      "favoriteFood": "pushpops"
    },
    "firstName": "ernie",
    "lastName": "duncan",
    "age": 29
  },
];

// function to retrieve the key value recursively
function valueByKey(obj, key) {
  let value = obj[key];
  if (value) {
    return value;
  }
  Object.keys(obj).forEach(function(k) {
    if (typeof obj[k] == 'object') {
      value = valueByKey(obj[k], key);
    }
  });
  return value;
}

const valuesNeeded = [{
    name: 'firstName'
  },
  {
    name: 'lastName'
  },
  {
    name: 'favoriteFood'
  }
];

// creating new object by passing the current object and key
let output = people.map(function(item) {
  let result = {};
  valuesNeeded.forEach(function(obj) {
    result[obj.name] = valueByKey(item, obj.name);
  });
  return result;
});

console.log(output);

Answer 2

您可以在顶层搜索所需的键，如果找不到，则转到子对象。这是我的方法：

function getValue(obj, key) {
    const keys = Object.keys(obj);
    // Check if key exist on this level
    if (keys.includes(key)) {
        return obj[key];
    } else {
        // else we try to find in child objects
        for (k of keys) {
            if (typeof obj[k] == 'object') {
                return getValue(obj[k], key);
            }
        }
    }
    //Return this if not found
    return '';
}

const people = [
    {
        foods: {
            favoriteFood: 'Ice cream',
        },
        firstName: 'John',
        lastName: 'Doe',
        age: 30,
    },
    {
        foods: {
            favoriteFood: 'pizza',
        },
        firstName: 'Jane',
        lastName: 'Doe',
        age: 39,
    },
    {
        foods: {
            favoriteFood: 'pbj',
        },
        firstName: 'Kevin',
        lastName: 'Baker',
        age: 22,
    },
    {
        foods: {
            favoriteFood: 'pushpops',
        },
        firstName: 'ernie',
        lastName: 'duncan',
        age: 29,
    },
];

const valuesNeeded = [
    { name: 'firstName' },
    { name: 'lastName' },
    { name: 'favoriteFood' },
];

let output = people.map((item) => {
    let result = {};
    valuesNeeded.forEach(function (obj) {
        result[obj.name] = getValue(item, obj.name);
    });
    return result;
});

console.log(output);

编辑1：您可以进一步将valuesNeeded简化为keysNeeded，并仅将键存储在数组中，如下所示keysNeeded = ['firatName', 'lastName', 'favoriteFood']。请相应地更改调用。