JS用新密钥创建一个新的对象数组

Question

所以我有两个JSON对象，房间和预留数组。 我想输出一个新的对象数组，这些对象包含来自这两个数组的数据和新的数据类型。 不确定如何创建新的数据类型。

var rooms = [
     {id:1, room:'treehouse'},
     {id:2, room:'casa'},
     {id:3, room:'vacation'},
     {id:4, room:'presidential'}
];

var reservations = [
  {id:1, roomID:'2', time:'2pm', location:'rome'},
  {id:2, roomID:'3', time:'4pm', location:'paris'},
  {id:3, roomID:'1', time:'4pm', location:'london'},
  {id:4, roomID:'2', time:'7pm', location:'rome'},
  {id:5, roomID:'1', time:'12pm', location:'london'},
  {id:6, roomID:'4', time:'4pm', location:'berlin'}
];

期望的输出：

var bookings = [
  {id: 1, roomid:1, time:'12pm',location:'london', roomname:'treehouse'},
  {id: 2, roomid:1, time:'4pm', location:'london', roomname:'treehouse'},
  {id: 3, roomid:2, time:'2pm', location:'rome', roomname:'casa'},
  {id: 4, roomid:2, time:'7pm', location:'rome', roomname:'casa'},
  {id: 5, roomid:3, time:'4pm', location:'paris', roomname:'vacation'},
  {id: 6, roomid:4, time:'4pm', location:'berlin', roomname:'presidential'}
]

我对如何做到这一点的逻辑感到困惑。 我正在考虑迭代reservations数组，并为每个预订抓住roomId并检查房间的地图结构然后输出。 我不太清楚该怎么做。

Answer 1

 var rooms = [ {id:1, room:'treehouse'}, {id:2, room:'casa'}, {id:3, room:'vacation'}, {id:4, room:'presidential'} ]; var reservations = [ {id:1, roomID:'2', time:'2pm', location:'rome'}, {id:2, roomID:'3', time:'4pm', location:'paris'}, {id:3, roomID:'1', time:'4pm', location:'london'}, {id:4, roomID:'2', time:'7pm', location:'rome'}, {id:5, roomID:'1', time:'12pm', location:'london'}, {id:6, roomID:'4', time:'4pm', location:'berlin'} ]; console.log( reservations.map(a=>({...a, roomname: rooms.find(b=>b.id==a.roomID).room})) ) 

Answer 2

您可以使用Array.reduce()创建具有房间名称的id地图的地图，然后在预订时使用Array.map()将具有给定id的房间与roomname合并：

 var rooms = [ {id:1, room:'treehouse'}, {id:2, room:'casa'}, {id:3, room:'vacation'}, {id:4, room:'presidential'} ]; var reservations = [ {id:1, roomID:'2', time:'2pm', location:'rome'}, {id:2, roomID:'3', time:'4pm', location:'paris'}, {id:3, roomID:'1', time:'4pm', location:'london'}, {id:4, roomID:'2', time:'7pm', location:'rome'}, {id:5, roomID:'1', time:'12pm', location:'london'}, {id:6, roomID:'4', time:'4pm', location:'berlin'} ]; let roomMap = rooms.reduce((a,curr)=>{ a[curr.id] = {roomname : curr.room}; return a; },{}); let result = reservations.map((o)=> Object.assign({},o,roomMap[o.roomID])); console.log(result); 

Answer 3

您可以记录Map的强大功能并获取新对象的附加信息。

 var rooms = [{ id: 1, room: 'treehouse' }, { id: 2, room: 'casa' }, { id: 3, room: 'vacation' }, { id: 4, room: 'presidential' }], roomMap = new Map(rooms.map(({ id: roomid, room: roomname }) => [roomid, { roomname }])), reservations = [{ id: 1, roomID: '2', time: '2pm', location: 'rome' }, { id: 2, roomID: '3', time: '4pm', location: 'paris' }, { id: 3, roomID: '1', time: '4pm', location: 'london' }, { id: 4, roomID: '2', time: '7pm', location: 'rome' }, { id: 5, roomID: '1', time: '12pm', location: 'london' }, { id: 6, roomID: '4', time: '4pm', location: 'berlin' }], bookings = reservations.map(o => Object.assign({}, o, roomMap.get(+o.roomID))); console.log(bookings); 

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Answer 4

 var reservations = [ {id:1, roomID: 2, time:'2pm', location:'rome'}, {id:2, roomID: 3, time:'4pm', location:'paris'}, {id:3, roomID: 1, time:'4pm', location:'london'}, {id:4, roomID: 2, time:'7pm', location:'rome'}, {id:5, roomID: 1, time:'12pm', location:'london'}, {id:6, roomID: 4, time:'4pm', location:'berlin'} ]; var rooms = [ {id:1, room:'treehouse'}, {id:2, room:'casa'}, {id:3, room:'vacation'}, {id:4, room:'presidential'} ]; reservations.forEach(reservation => { reservation.roomname = rooms.find(({id}) => id === reservation.roomID).room; }) console.log(reservations); 

Answer 5

你可以用

reservations.map(function (val) {
    return Object.assign(val, rooms.find(function (item) {
        return val.roomID === item.id ? item : {}
    }))
})

如果这对您有用，请告诉我。

Answer 6

您可以使用Array.prototype.reduce和Array.prototype.map创建查找，以迭代reservations数组并构造所需的输出：

 var rooms = [{id:1, room:'treehouse'},{id:2, room:'casa'},{id:3, room:'vacation'},{id:4, room:'presidential'}]; var reservations = [{id:1, roomID:'2', time:'2pm', location:'rome'},{id:2, roomID:'3', time:'4pm', location:'paris'},{id:3, roomID:'1', time:'4pm', location:'london'},{id:4, roomID:'2', time:'7pm', location:'rome'},{id:5, roomID:'1', time:'12pm', location:'london'},{id:6, roomID:'4', time:'4pm', location:'berlin'}]; var roomHash = rooms.reduce((all, {id, room}) => ({...all, [id]: room }), {}); var result = reservations.map(item => ({...item, room: roomHash[item.roomID]})); console.log(result);