如何使用JS reduce将两个objets数组合并到一个对象数组中，其中一个特定的键/值匹配

Question

我试图了解如何将两个对象数组合并为一个对象数组，并将两个数组中的特定键值匹配到一起

我有一个例子，说明我的目标是什么，但没有一个片段，因为我不知道如何开始和做它

两个对象

const objA = [
  {
    "id": "DEMO-1",
    "siteId": "01",
  },
  {
    "id": "DEMO-2",
    "siteId": "02",
  },
  {
    "id": "DEMO-3",
    "siteId": "05",
  },
  {
    "id": "DEMO-4",
    "siteId": null,
  },
  {
    "id": "DEMO-5",
    "siteId": null,
  },
  {
    "id": "DEMO-6",
    "siteId": "05",
  },
  {
    "id": "DEMO-7",
    "siteId": "02",
  },
];

const objB = [
  { siteId: '01', principalInvestigator: { name: 'Jake' } },
  { siteId: '02', principalInvestigator: { name: 'Claudia' } },
  { siteId: '05', principalInvestigator: null },
];

当我们在同一个siteId上有匹配时，这两个对象必须组合在一起。

结果应该是一个新的对象数组，情况如下：

我们有一个匹配的objB.principalInvestigator

• there，结果是objA +是不匹配的，然后添加principalInvestigator: null

const result = [
  {
    "id": "DEMO-1",
    "siteId": "01",
    "principalInvestigator": { name: 'Jake' }
  },
  {
    "id": "DEMO-2",
    "siteId": "02",
    "principalInvestigator": { name: 'Claudia' }
  },
  {
    "id": "DEMO-3",
    "siteId": "05",
    "principalInvestigator": null
  },
  {
    "id": "DEMO-4",
    "siteId": null,
    "principalInvestigator": null
  },
  {
    "id": "DEMO-5",
    "siteId": null,
    "principalInvestigator": null
  },
  {
    "id": "DEMO-6",
    "siteId": "05",
    "principalInvestigator": null
  },
  {
    "id": "DEMO-7",
    "siteId": "02",
    "principalInvestigator": { name: 'Claudia' }
  },
]

Answer 1

您可以为第二个对象(按siteId键)创建一个查找映射，然后将第一个数组映射到扩展对象，以便从映射中查找principalInvestigator数据：

​

const objA = [{"id": "DEMO-1","siteId": "01",},{"id": "DEMO-2","siteId": "02",},{"id": "DEMO-3","siteId": "05",},{"id": "DEMO-4","siteId": null,},{"id": "DEMO-5","siteId": null,},{"id": "DEMO-6","siteId": "05",},{"id": "DEMO-7","siteId": "02",},];

const objB = [
  { siteId: '01', principalInvestigator: { name: 'Jake' } },
  { siteId: '02', principalInvestigator: { name: 'Claudia' } },
  { siteId: '05', principalInvestigator: null },
];

const map = new Map(objB.map(o => [o.siteId, o.principalInvestigator]));
const result = objA.map(o => ({...o, prinicalInvestigator: map.get(o.siteId) ?? null}));
console.log(result);

​

注意，如果第一个数组中的多个对象具有相同的siteId，那么它们的新prinicalInvestigator属性将引用同一个对象。

Answer 2

对于objA中的每个objA，使用find在objB中定位相应的对象

result = objA.map(a => ({...a,
    ...(objB.find(b => b.siteId === a.siteId) ?? {principalInvestigator: null})
}))