更新数组中的对象（按值匹配）

Question

有第一个数组（a）和第二个更新值（b）。需要通过 IDs 对象更新匹配并获取结果数组 (c)。我怎样才能让它变得简单快捷？

let a = [
    { id: 1, activated: '0' },
    { id: 2, activated: '0' },
    { id: 3, activated: '0' },
    { id: 4, activated: '0' },
  ]
let b = [
    { id: 2, activated: '1' },
    { id: 3, activated: '1' },
  ]
//Result array:
c = [ 
  { id: 1, activated: '0' },
  { id: 2, activated: '1' },
  { id: 3, activated: '1' },
  { id: 4, activated: '0' },
]

Answer 1

我建议您创建一个对象，其中键是快速访问的 id。

使用新创建的对象，您可以使用id 作为键从b 访问特定对象并更新对象。

最后，使用函数Object.values得到数组c。

这是假设 b 的 id 存在于 a 中。

let a = [    { id: 1, activated: '0' },    { id: 2, activated: '0' },    { id: 3, activated: '0' },    { id: 4, activated: '0' }]
let b = [    { id: 2, activated: '1' },    { id: 3, activated: '1' }];
let newA = a.reduce((a, {id, ...rest}) => ({...a, ...{[id]: {id, ...rest}}}), {});

b.forEach(({id, activated}) => newA[id].activated = activated);

let c = Object.values(newA);
console.log(c);

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Answer 2

你可以使用Array.map，然后foreach对象，你可以使用Array.find通过ID检查它是否存在于arrayB中，如果存在则返回。

let a = [
    { id: 1, activated: '0' },
    { id: 2, activated: '0' },
    { id: 3, activated: '0' },
    { id: 4, activated: '0' },
];

let b = [
    { id: 2, activated: '1' },
    { id: 3, activated: '1' },
];

const c = a.map(e => {
    let newValue = b.find(n => e.id === n.id);

    if (!!newValue) {
        return newValue;
    }

    return e;
});

Answer 3

这就是我的处理方式：

function mergeArray(...toMerge) {
  let output = {};
  toMerge.forEach(arr => {
    arr.forEach(item => {
      output[item.id] = item;
    });
  });
  return Object.values(output);
}

let a = [
    { id: 1, activated: '0' },
    { id: 2, activated: '0' },
    { id: 3, activated: '0' },
    { id: 4, activated: '0' },
  ];
let b = [
    { id: 2, activated: '1' },
    { id: 3, activated: '1' },
  ];

console.dir(mergeArray(a, b)) // can merge n-number of arrays.

Answer 4

let c = a.map( aObj => {
  let found = b.find(bObj => bObj.id === aObj.id);
    if(found) return found;
    else return aObj;
});

Answer 5

您可以创建第二个数组的映射，以id 为键，然后当您将a 对象复制到c 中时，使用该映射将匹配的b 对象合并到它们中。

let a = [{ id: 1, activated: '0' },{ id: 2, activated: '0' },{ id: 3, activated: '0' },{ id: 4, activated: '0' }]
let b = [{ id: 2, activated: '1' },{ id: 3, activated: '1' }];

let map = new Map(b.map(o => [o.id, o]));
let c = a.map(o => ({...o, ...map.get(o.id)}));

console.log(c);

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Answer 6

从数组 a 的 id 构建散列。通过在哈希中查找索引 b.id 来迭代数组 b，如果在 a 中找到实际大小，则该属性已激活。否则（在示例 id: 5 中）在旧数组 a 中不存在此 id 的条目，因此创建一个新条目并实现哈希。

let a = [
    { id: 1, activated: '0' },
    { id: 2, activated: '0' },
    { id: 3, activated: '0' },
    { id: 4, activated: '0' },
  ];
let b = [
    { id: 2, activated: '1' },
    { id: 3, activated: '1' },
    { id: 5, activated: '0' },
  ];

let hash = a.flatMap(el => el.id);

b.forEach(elem => {
    index = hash.indexOf(elem.id);
    if (index!=-1) 
        a[index].activated=elem.activated;
     else {
        a.push(elem);
        hash.push(elem.id);
     }
});
  
console.log(a);  

Answer 7

试试这个

let a = [{
    id: 1,
    activated: '0'
  },
  {
    id: 2,
    activated: '0'
  },
  {
    id: 3,
    activated: '0'
  },
  {
    id: 4,
    activated: '0'
  },
]
let b = [{
    id: 2,
    activated: '1'
  },
  {
    id: 3,
    activated: '1'
  },
]
//Result array:
c = [{
    id: 1,
    activated: '0'
  },
  {
    id: 2,
    activated: '1'
  },
  {
    id: 3,
    activated: '1'
  },
  {
    id: 4,
    activated: '0'
  },
]

let result = a.map(e => {
  let p = b.find(be => be.id === e.id)
  if (p) e.activated = p.activated;
  return e
})

console.log(result)