将所有客房的TimeSlots组合起来

Question

假设我有三个房间，分别有可用的时隙和总时隙。

总时隙：

["09:00-10:00","10:00-11:00", "11:00-12:00",  "12:00-13:00", "13:00-14:00", "14:00-15:00"]

Room1可用于

[ "10:00-11:00", "11:00-12:00",  "12:00-13:00"]

Room2可用于

[ "11:00-12:00",  "12:00-13:00", "13:00-14:00"]

Room3可用于

[ "12:00-13:00", "13:00-14:00", "14:00-15:00"]

我想用房间号过滤掉可用的时隙，所以我希望我的输出是这样的。

[{
    slot: "09:00-10:00",
    available: false,
    space : []
},{
    slot: "10:00-11:00",
    available: true,
    space : ["room1"]
},{
    slot: "11:00-12:00",
    available: true,
    space : ["room1", "room2"]
},{
    slot: "12:00-13:00",
    available: true,
    space : ["room1", "room2", "room3"]
},{
    slot: "13:00-14:00",
    available: true,
    space : ["room2", "room3"]
},{
    slot: "14:00-15:00",
    available: true,
    space : ["room3"]
}]

我试过的是：

​

const ts = ['10:00-10:30', '10:30-11:00', '11:00-11:30', '11:30-12:00', '12:00-12:30', '12:30-13:00', '13:00-13:30', '13:30-14:00', '14:00-14:30', '14:30-15:00', '15:00-15:30', '15:30-16:00'];

const room1 = ["11:00-11:30", "13:05-13:35", "14:05-14:15"];

const avail = (ts, booked) =>
  ts.map(item => {
    const [start, end] = item.split('-');
    const isBooked = booked
      .map(item => item.split('-'))
      .some(([bookedStart, bookedEnd]) =>
        (start >= bookedStart && start < bookedEnd) ||
        (end > bookedStart && end <= bookedEnd) ||
        (bookedStart >= start && bookedStart < end));
    return {
      slot: `${start}-${end}`,
      isBooked
    };
  })



console.log(avail(ts, room1));

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  min-height: 100%
}

​

Answer 1

我已尽力使它行得通。请检查这是否符合您的要求。

​

const ts = ['10:00-10:30','10:30-11:00','11:00-11:30','11:30-12:00','12:00-12:30','12:30-13:00','13:00-13:30','13:30-14:00','14:00-14:30','14:30-15:00','15:00-15:30','15:30-16:00'];

let rooms = {"room100":["11:00-11:30", "13:05-13:35", "14:05-14:15"],"room200":[ "11:00-12:00",  "12:00-13:00", "13:00-14:00"],"room300":[ "12:00-13:00", "13:00-14:00", "14:00-15:00"]};
var final = [];
const avail = (ts, room) =>
	ts.map(item => {
		const[start, end] = item.split('-');
    let slot = [];
    
  for(var key in room)
        {
    	let avilable = room[key]
			.map(item => item.split('-'))
			.some(([bookedStart, bookedEnd]) =>
				(start >= bookedStart && start < bookedEnd) ||
				(end > bookedStart && end <= bookedEnd) ||
				(bookedStart >= start && bookedStart < end));
       slot = avilable == false ? [...slot] : [...slot,key];
       console.log()
    }
    const isBooked = slot.length > 0 ? true :false;
    final.push({"slot":`${start}-${end}`, "available": isBooked, "space": slot })
    //console.log(`${start}-${end}`,isBooked, slot);
	})
  
avail(ts,rooms);

console.log(final)

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​

Answer 2

您可以使用filter和Object.entries (假设您有一个rooms对象)来完成这个任务。这样做的想法是，对于ts数组中的每一个时隙，您都希望找到(filter)具有该时隙中的可用性的房间。

我还建议将检查某个时间是否与某个时隙重叠的逻辑抽象为另一个函数，以便您可以单独测试它：

​

const ts = ['10:00-10:30','10:30-11:00','11:00-11:30','11:30-12:00','12:00-12:30','12:30-13:00','13:00-13:30','13:30-14:00','14:00-14:30','14:30-15:00','15:00-15:30','15:30-16:00'];
const rooms = {
  'room1': ["11:00-11:30", "13:05-13:35", "14:05-14:15"],
  'room2': [ "11:00-12:00",  "12:00-13:00", "13:00-14:00"],
  'room3': [ "12:00-13:00", "13:00-14:00", "14:00-15:00"]
};

const overlaps = (slot, booking) => {
  const [start, end] = slot.split('-');
  const [bookedStart, bookedEnd] = booking.split('-');
  return (start >= bookedStart && start < bookedEnd) ||
		 (end > bookedStart && end <= bookedEnd) ||
		 (bookedStart >= start && bookedStart < end);
};

const res = ts.map(slot => {
  const space = Object.entries(rooms)
    .filter(([_, availableSlots]) => availableSlots.some(roomSlot => overlaps(slot, roomSlot)))
    .map(([room, _]) => room);
  return {slot, space, available: space.length > 0};
});

console.log(res);

​

Answer 3

您可以使用一些嵌套的迭代结构，并检查想要的插槽中是否有房间可用。

​

var slots = [ '10:00-11:00',
  '10:15-11:15',
  '10:30-11:30',
  '10:45-11:45',
  '11:00-12:00' ],
    rooms = { '5d396fd3f0315c3f56b255d4-5d396fe2f0315c3f56b25a79': [],
  '5d396fa0f0315c3f56b245d6': [ '11:00-12:00' ] },
    roomEntries = Object.entries(rooms),
    result = slots.map(slot => {
        var [sStart, sEnd] = slot.split('-'),
            space = roomEntries
                .filter(([room, a]) => a.some(available => {
                    var [aStart, aEnd] = available.split('-');
                    return (sStart >= aStart && sEnd <= aEnd || sStart <= aStart && sEnd > aStart || sStart < aEnd && sEnd >= aEnd);
                }))
                .map(([k]) => k);
        return { slot, available: !!space.length, space };
    });

console.log(result);

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​