我有一个 Javascript 对象数组，如下所示。[ { email: 'alex@test.com', fn: 'Alex', sn: 'McPherson', phone: '01233xxxxx', hours: '40', rate: '20', amount: '200', vat: '60', agency: 'test', start: '08/06/2017', end: '10/06/2017', shipping: { addresses: [ { id: '1234', area: 'xzy' }, { id: '2345', area: 'uhj' } ] } }, { email: 'mike@test.com', fn: 'Mike', sn: 'Mann', phone: '01233xxxxx', hours: '50', rate: '70', amount: '500', vat: '90', agency: 'test', start: '08/06/2017', end: '10/06/2017', shipping: { addresses: [ { id: '1234', area: 'xzy' }, { id: '3456', area: 'uio' } ] } }, { email: 'fred@test.com', fn: 'Fred', sn: 'Frogg', phone: '01233xxxxx', hours: '80', rate: '90', amount: '800', vat: '100', agency: 'test', start: '08/06/2017', end: '10/06/2017', shipping: { addresses: [ { id: '4567', area: 'asdaf' }, { id: '3456', area: 'uio' } ] } }, { email: 'alex@test.com', fn: 'Alex', sn: 'McPherson', phone: '01233xxxxx', hours: '90', rate: '30', amount: '900', vat: '120', agency: 'test', start: '08/06/2017', end: '10/06/2017', shipping: { addresses: [ { id: '4567', area: 'asdaf' }, { id: '5678', area: 'asdf' } ] } }]我理想的情况是将那些具有相同值（shipping.addresses.id）的对象分组到自己的对象子数组中。预期结果。我可以使用特定键（下面的代码）使用特定属性对输入数组进行分组，但我似乎无法根据数组本身的键来重新排序数组。Array.from( data.reduce( (acc, o) => (acc.get(o.email).push(o), acc), new Map(data.map( o => [o.email, []] )) ), ([key, value]) => value)
1 回答
慕桂英3389331
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您可以data使用 as 键将数组缩减为一个对象shipping.addresses.id，并使用 返回一个数组Object.values()。您将需要迭代addresses每个对象的数组，并在遇到每个对象时为每个对象创建一个条目id，并推送到这些条目以获取具有相同id.
const byAddressId = Object.values(
data.reduce((a, o) => {
o.shipping.addresses.forEach(({id, area}) => {
a[id] = {...a[id] ?? {id: id, area: area, data: []}};
a[id]['data'].push({...o});
});
return a;
}, {}));
const data = [{"email": "alex@test.com","fn": "Alex","sn": "McPherson","phone": "01233xxxxx","hours": "40","rate": "20","amount": "200","vat": "60","agency": "test","start": "08/06/2017","end": "10/06/2017","shipping": { "addresses": [ { "id": "1234", "area": "xzy" }, { "id": "2345", "area": "uhj" } ]}},{"email": "mike@test.com","fn": "Mike","sn": "Mann","phone": "01233xxxxx","hours": "50","rate": "70","amount": "500","vat": "90","agency": "test","start": "08/06/2017","end": "10/06/2017","shipping": { "addresses": [ { "id": "1234", "area": "xzy" }, { "id": "3456", "area": "uio" } ]}},{"email": "fred@test.com","fn": "Fred","sn": "Frogg","phone": "01233xxxxx","hours": "80","rate": "90","amount": "800","vat": "100","agency": "test","start": "08/06/2017","end": "10/06/2017","shipping": { "addresses": [ { "id": "4567", "area": "asdaf" }, { "id": "3456", "area": "uio" } ]}},{"email": "alex@test.com","fn": "Alex","sn": "McPherson","phone": "01233xxxxx","hours": "90","rate": "30","amount": "900","vat": "120","agency": "test","start": "08/06/2017","end": "10/06/2017","shipping": { "addresses": [ { "id": "4567", "area": "asdaf" }, { "id": "5678", "area": "asdf" } ]}}];
// return array of Object.values from the accumulator
const byAddressId = Object.values(
// reduce the data array into an object with shipping.addresses.id as keys
data.reduce((a, o) => {
// iterate over all addresses for each element
o.shipping.addresses.forEach(({id, area}) => {
// check if an id entry exists, otherwise create one
a[id] = {...a[id] ?? {id: id, area: area, data: []}};
// push the object to the data array of the id object
a[id]['data'].push({...o});
});
return a;
}, {}));
console.log(byAddressId);
话虽这么说，与问题中包含的示例map()相比，您可以使用相同的方法来节省两次调用。group by email
const byEmail = Object.values(
data.reduce((a, o) => (a[o.email] = [...a[o.email] ?? [], {...o}], a), {}));添加回答
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