我有一个对象数组，要根据特定条件进行分组。
下面是原始数组：
let meetingOriginal = [{
id: 19,
datetime: '2019-01-02T13:28:03+05:30',
status: 'proposed',
createdBy: 'buyer',
buyerId: 2,
buyerFirstName: 'Demo',
buyerLastName: 'User',
buyerProfileUrl: null,
petId: 1,
petName: 'Penny',
petMedia: 'https://s3.us-east-2.amazonaws.com/s3-petquest-dev/images/1546324490530.jpg',
ownerFirstName: 'Faizan',
ownerLastName: 'Saiyed',
ownerProfileUrl: null,
ownerId: 3,
},
{
id: 20,
datetime: '2019-01-03T13:28:06+05:30',
status: 'proposed',
createdBy: 'buyer',
buyerId: 2,
buyerFirstName: 'Demo',
buyerLastName: 'User',
buyerProfileUrl: null,
petId: 1,
petName: 'Penny',
petMedia: 'https://s3.us-east-2.amazonaws.com/s3-petquest-dev/images/1546324490530.jpg',
ownerFirstName: 'Faizan',
ownerLastName: 'Saiyed',
ownerProfileUrl: null,
ownerId: 3,
},
{
id: 21,
datetime: '2019-01-04T13:28:09+05:30',
status: 'proposed',
createdBy: 'buyer',
buyerId: 2,
buyerFirstName: 'Demo',
buyerLastName: 'User',
buyerProfileUrl: null,
petId: 1,
petName: 'Penny',
petMedia: 'https://s3.us-east-2.amazonaws.com/s3-petquest-dev/images/1546324490530.jpg',
ownerFirstName: 'Faizan',
ownerLastName: 'Saiyed',
ownerProfileUrl: null,
ownerId: 3,
},
{
id: 22,
datetime: '2019-01-04T13:28:09+05:30',
status: 'proposed',
createdBy: 'buyer',
buyerId: 3,
buyerFirstName: 'Demo',
buyerLastName: 'User',
buyerProfileUrl: null,
petId: 2,
petName: 'Jerry',
petMedia: 'https://static1.squarespace.com/static/52784cdde4b07cdbb003018f/t/56259d40e4b043d43a71cc73/1445305665196/Cavapoo-dog-painting-square-a.jpg?format=500w',
ownerFirstName: 'Jony',
ownerLastName: 'Snowy',
ownerProfileUrl: null,
ownerId: 5,
}
];
如何仅使用.map（）或.filter（）方法（严格来说，没有库或for＆foreach函数）对这些对象进行分组，如下所示？
let requiredMeeting = [{
buyerId: 2,
buyerFirstName: 'Demo',
buyerLastName: 'User',
buyerProfileUrl: null,
petId: 1,
petName: 'Penny',
petMedia: 'https://s3.us-east-2.amazonaws.com/s3-petquest-dev/images/1546324490530.jpg',
ownerFirstName: 'Faizan',
ownerLastName: 'Saiyed',
ownerId: 3,
ownerProfileUrl: 'https://s3.us-east-2.amazonaws.com/s3-petquest-dev/images/1546338985171.jpg',
meetings: [{
id: 19,
datetime: '2019-01-02T13:28:03+05:30',
status: 'proposed',
createdBy: 'buyer'
},
{
id: 20,
datetime: '2019-01-03T13:28:06+05:30',
status: 'accepted',
createdBy: 'buyer'
},
{
id: 21,
datetime: '2019-01-04T13:28:09+05:30',
status: 'proposed',
createdBy: 'buyer'
}
]
},
{
buyerId: 3,
buyerFirstName: 'Demo',
buyerLastName: 'User',
buyerProfileUrl: null,
petId: 2,
petName: 'Jerry',
petMedia: 'https://static1.squarespace.com/static/52784cdde4b07cdbb003018f/t/56259d40e4b043d43a71cc73/1445305665196/Cavapoo-dog-painting-square-a.jpg?format=500w',
ownerFirstName: 'Jony',
ownerLastName: 'Snowy',
ownerId: 5,
ownerProfileUrl: 'https://cdna.artstation.com/p/assets/images/images/002/898/132/large/hector-jenz-jonhsnow-square-detaill.jpg?1467032227',
meetings: [{
id: 22,
datetime: '2019-01-04T13:28:09+05:30',
status: 'proposed',
createdBy: 'buyer'
}]
}
];
可能有多个对象以及不同的petId和ownerId。
这是小提琴： LINK
答案 0 :(得分：3)
您可以使用Reduce执行此操作
let meetingOriginal =[{id:19,datetime:'2019-01-02T13:28:03+05:30',status:'proposed',createdBy:'buyer',buyerId:2,buyerFirstName:'Demo',buyerLastName:'User',buyerProfileUrl:null,petId:1,petName: 'Penny',petMedia: 'https://s3.us-east-2.amazonaws.com/s3-petquest-dev/images/1546324490530.jpg',ownerFirstName:'Faizan',ownerLastName: 'Saiyed',ownerProfileUrl: null,ownerId: 3,},{id: 20,datetime: '2019-01-03T13:28:06+05:30',status: 'proposed', createdBy: 'buyer', buyerId: 2, buyerFirstName: 'Demo', buyerLastName: 'User', buyerProfileUrl: null, petId: 1, petName: 'Penny', petMedia: 'https://s3.us-east-2.amazonaws.com/s3-petquest-dev/images/1546324490530.jpg', ownerFirstName: 'Faizan', ownerLastName: 'Saiyed', ownerProfileUrl: null, ownerId: 3,},{ id: 21, datetime: '2019-01-04T13:28:09+05:30', status: 'proposed', createdBy: 'buyer', buyerId: 2, buyerFirstName: 'Demo', buyerLastName: 'User', buyerProfileUrl: null, petId: 1, petName: 'Penny', petMedia: 'https://s3.us-east-2.amazonaws.com/s3-petquest-dev/images/1546324490530.jpg', ownerFirstName: 'Faizan', ownerLastName: 'Saiyed', ownerProfileUrl: null, ownerId: 3,},{ id: 22, datetime: '2019-01-04T13:28:09+05:30', status: 'proposed', createdBy: 'buyer', buyerId: 3, buyerFirstName: 'Demo', buyerLastName: 'User', buyerProfileUrl: null, petId: 2, petName: 'Jerry', petMedia: 'https://static1.squarespace.com/static/52784cdde4b07cdbb003018f/t/56259d40e4b043d43a71cc73/1445305665196/Cavapoo-dog-painting-square-a.jpg?format=500w', ownerFirstName: 'Jony', ownerLastName: 'Snowy', ownerProfileUrl: null, ownerId: 5,}];
let op = meetingOriginal.reduce((o,c)=>{
if(o[c.buyerId]){
o[c.buyerId]['meetings'].push({
id: c.id, datetime: c.dateTime, status: c.status, createdBy: c.createdBy
})
} else {
let { buyerId,buyerFirstName,buyerLastName,buyerProfileUrl,petId,petName,petMedia,ownerFirstName,ownerLastName,ownerId,ownerProfileUrl} = c
o[c.buyerId]={buyerId: buyerId, buyerFirstName:buyerFirstName, buyerLastName:buyerLastName, buyerProfileUrl:buyerProfileUrl, petId:petId, petName:petName, petMedia:petMedia, ownerFirstName:ownerFirstName, ownerLastName:ownerLastName, ownerId:ownerId, ownerProfileUrl:ownerProfileUrl, meetings : [{ datetime: c.dateTime, status: c.status, createdBy: c.createdBy }]}}
return o;
},{})
let finalOp = Object.values(op);
console.log(finalOp)
答案 1 :(得分：1)
我已经使用map()尝试了解决方案：
let inputObjs = [
{
grpId: 2,
name: 'group_2',
id: 1,
date: '2019-01-02'
},
{
grpId: 2,
name: 'group_2',
id: 2,
date: '2019-01-03'
},
{
grpId: 5,
name: 'group_5',
id: 1,
date: '2019-01-01'
}
];
var map = new Map();
inputObjs.map(obj => {
var grp = obj.grpId;
var mapVal = map.get(grp);
if (! mapVal) {
map.set(grp,
{
grpId : obj.grpId,
name : obj.name,
meetings : [
{id: obj.id, date: obj.date}
]
}
);
}
else {
mapVal.meetings.push({id: obj.id, date: obj.date});
}
});
// print the output
for (let e of map.values()) {
console.log(e);
}
结果：
[
{
grpId: 2,
name: 'group_2',
meetings: [
{id: 1, date: '2019-01-02'},
{id: 2, date: '2019-01-03'}
]
},
{
grpId: 5,
name: 'group_5',
meetings: [
{id: 1, date: '2019-01-01'}
]
}
];