如果具有相同的鍵值，則將鍵值添加到對象數組

Question

我正在嘗試根據另一個數組中的相應值向一個數組中的所有對象添加一個具有特定值的屬性。

const array1 = [
  {
    id: 1,
    date: '2022.05.01',
    name: 'john'
  }, {
    id: 2,
    date: '2022.05.01',
    name: 'sam'
  }, {
    id: 3,
    date: '2022.05.03',
    name: 'john'
  }, {
    id: 4,
    date: '2022.05.06',
    name: 'jack'
  },
 ]
    

此數組包含需要進行的必要修改：

const array2 = [
  {
    name: 'john',
    isCanceled: true,
  }, {
    name: 'jack',
    isCanceled: false,
  }, {
    name: 'sam',
    isCanceled: false,
  },
 ]

如果array1中對象的名稱是john ，則isCanceled應設置為true ，但如果是jack或sam則應設置為false ，如下所示：

const resultArray = [
  {
    id: 1,
    date: '2022.05.01',
    name: 'john',
    isCanceled: true,
  }, {
    id: 2,
    date: '2022.05.01',
    name: 'sam'
    isCanceled: false,
  }, {
    id: 3,
    date: '2022.05.03',
    name: 'john'
    isCanceled: true,
  }, {
    id: 4,
    date: '2022.05.06',
    name: 'jack'
    isCanceled: false,
  },
 ];

Answer 1

構建一個包含狀態的status查找對象，並映射您的數組以包含查找對象中的匹配值：

const status = Object.fromEntries(
  array2.map(({name, isCanceled}) => [name, isCanceled])
);

const resultArray = array1.map(({id, date, name}) => ({
  id,
  date,
  name,
  isCanceled: status[name]
}));

或者，如果您只想修改array1而不是創建新數組，則可以將第二步替換為：

array1.forEach(v => v.isCanceled = status[v.name]);

---

完整片段：

 const array1 = [{ id: 1, date: '2022.05.01', name: 'john' }, { id: 2, date: '2022.05.01', name: 'sam' }, { id: 3, date: '2022.05.03', name: 'john' }, { id: 4, date: '2022.05.06', name: 'jack' }, ]; const array2 = [{ name: 'john', isCanceled: true, }, { name: 'jack', isCanceled: false, }, { name: 'sam', isCanceled: false, }, ]; const status = Object.fromEntries( array2.map(({name, isCanceled}) => [name, isCanceled]) ); const resultArray = array1.map(({id, date, name}) => ({ id, date, name, isCanceled: status[name] })); console.log(resultArray);

您是否想知道為什么我更喜歡構建查找對象而不是在map()操作中使用find() ：就復雜性而言，查找表的復雜度為O(1) ，而find()方法為O( n) .

Answer 2

你可以做一些事情：

const result = array1.map(person => {
  const lookPerson = array2.find(person2 => person2.name === person.name);
  return (lookPerson) ? { 
    ...person,
    isCanceled: lookPerson.isCanceled
  } : person;
});

• 結合 array.map 來變換 array1

• 使用 array.find 在 array1 和 array2 之間獲取相似人的第一次迭代

• 合並對象

  { ...person, isCanceled: lookPerson.isCanceled } 

 const array1 = [{ id: 1, date: '2022.05.01', name: 'john' }, { id: 2, date: '2022.05.01', name: 'sam' }, { id: 3, date: '2022.05.03', name: 'john' }, { id: 4, date: '2022.05.06', name: 'jack' }, ]; const array2 = [{ name: 'john', isCanceled: true, }, { name: 'jack', isCanceled: false, }, { name: 'sam', isCanceled: false, }, ]; const result = array1.map(person => { const lookPerson = array2.find(person2 => person2.name === person.name); return (lookPerson) ? { ...person, isCanceled: lookPerson.isCanceled } : person; }); console.log(result);

Answer 3

您可以更簡單地做到這一點，首先將您的array2轉換為一個對象，然后在map中使用查找，如下所示：

 const array1 = [{id:1,date:"2022.05.01",name:"john"},{id:2,date:"2022.05.01",name:"sam"},{id:3,date:"2022.05.03",name:"john"},{id:4,date:"2022.05.06",name:"jack"},]; const array2 = [{name:"john",isCanceled:true},{name:"jack",isCanceled:false},{name:"sam",isCanceled:false}]; const arrCancel = array2.reduce((a, { name, isCanceled }) => { a[name] = isCanceled; return a; }, {}); const resultArr = array1.map(e => { e.isCanceled = arrCancel[e.name]; return e; }); console.log(resultArr);

 .as-console-wrapper { max-height: 100% !important; top: auto; }

Answer 4

你可以做：

 const array1 = [{id: 1,date: '2022.05.01',name: 'john'}, {id: 2,date: '2022.05.01',name: 'sam'}, {id: 3,date: '2022.05.03',name: 'john'}, {id: 4,date: '2022.05.06',name: 'jack'}] const array2 = [{name: 'john',isCanceled: true,}, {name: 'jack',isCanceled: false,}, {name: 'sam',isCanceled: false,}] const array2Hash = array2.reduce((a, { name: n, isCanceled: ic}) => (a[n] = ic, a), {}) const result = array1.map(u => ({ ...u, isCanceled: array2Hash[u.name] })) console.log(result)

Answer 5

resultArray 是array1的映射，加上基於array2的isCanceled ：

resultArray = array1.map( a => Object.assign(
  {},
  a,
  {isCanceled: array2.find(b => a.name == b.name)?.isCanceled||false}) )