将对象数组变成对象

Question

我有一个分支数组，看起来像这样：

let branches = [
  {
    id: 21,
    name: "Branch 1",
    opening_times: [ {}, {}, {} ] // Array of objects (Monday, Tuesday etc)
  },
  {
    id: 22,
    name "Branch 2"
    opening_times: [ {}, {}, {} ] // Array of objects (Monday, Tuesday etc)
  },
  // .. etc
]

但我想将其变成一个对象，每个对象的名称均作为关键字。

所需的输出：

branches = {
  "Branch 1": {
    id: 21,
    opening_times: [ {}, {}, {} ] // Array of objects (Monday, Tuesday etc)
  },
  "Branch 2": {
    id: 22,
    opening_times: [ {}, {}, {} ] // Array of objects (Monday, Tuesday etc)
  }
}

尝试过：

branches = {
  "Branch 1": {
    id: 21,
    opening_times: [ {}, {}, {} ] // Array of objects (Monday, Tuesday etc)
  },
  "Branch 2": {
    id: 22,
    opening_times: [ {}, {}, {} ] // Array of objects (Monday, Tuesday etc)
  }
}

但是当然映射给了我一个数组输出：

let newBranches = branches.map(branch => (
  {
    [branch.name]: {
      id: branch.id,
      days: branch.opening_times
    }
  }
));
console.log(newBranches)

任何人都可以帮我指出正确的方向，以将name密钥作为对象本身来获得一个新对象吗？

Answer 1

我只是使用一个简单的for-of循环。您会得到reduce答案，但是reduce这里所做的只是增加了复杂性。

const result = {};
for (const {name, id, opening_times} of branches) {
  result[name] = {id, opening_times};
}

现场示例：

const result = {};
for (const {name, id, opening_times} of branches) {
  result[name] = {id, opening_times};
}

let branches = [
  {
    id: 21,
    name: "Branch 1",
    opening_times: [ {}, {}, {} ] // Array of objects (Monday, Tuesday etc)
  },
  {
    id: 22,
    name: "Branch 2",
    opening_times: [ {}, {}, {} ] // Array of objects (Monday, Tuesday etc)
  },
  // .. etc
];
const result = {};
for (const {name, id, opening_times} of branches) {
  result[name] = {id, opening_times};
}
console.log(result);

补充Code Maniac关于使用休息的建议：

const result = {};
for (const {name, ...entry} of branches) {
  result[name] = entry;
}

现场示例：

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

const result = {};
for (const {name, ...entry} of branches) {
  result[name] = entry;
}

两者略有不同，因为第一个显式仅使用id和opening_times结果，而其余版本使用以外的所有属性name。当然，可读性有所不同（显式与隐式），但是我会在每个地方都使用它们。

Answer 2

通过简单的reduce()操作和对象分解：

const branches = [{
    id: 21,
    name: "Branch 1",
    opening_times: []
  },
  {
    id: 22,
    name: "Branch 2",
    opening_times: []
  }
];

const result = branches.reduce((a, {name, ...v}) => (a[name] = v, a), {});

console.log(result);