我有一个如下的Javascript对象数组。
[
{
email: '[email protected]',
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: '[email protected]',
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: '[email protected]',
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: '[email protected]',
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。
[
{
id: '1234',
area: 'xzy',
data: [
{
email: '[email protected]',
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: '[email protected]',
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: 'uhj'
}
]
}
}
]
},
{
id: '2345',
area: 'uhj',
data: [
{
email: '[email protected]',
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: 'uio'
}
]
}
}
]
},
{
id: '3456',
area: 'uio',
data: [
{
email: '[email protected]',
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: '[email protected]',
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'
}
]
}
}
]
}
]
我可以使用特定的键（使用下面的代码）使用特定的属性来对输入数组进行分组，但是我似乎无法基于基于键（本身就是数组）来求助于该数组。
Array.from(
data.reduce(
(acc, o) => (acc.get(o.email).push(o), acc),
new Map(data.map( o => [o.email, []] ))
), ([key, value]) => value
)
您可以data使用shipping.addresses.id键将数组简化为对象，并使用来返回数组Object.values()。您将需要遍历addresses每个对象的数组，并id在遇到每个对象时为其创建一个条目，并为具有相同元素的后续元素推送到这些条目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": "[email protected]","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": "[email protected]","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": "[email protected]","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": "[email protected]","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);话虽如此，与您在问题中包含map()的group by email示例相比，您可以使用相同的方法来节省两个调用。
const byEmail = Object.values(
data.reduce((a, o) => (a[o.email] = [...a[o.email] ?? [], {...o}], a), {}));
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