Question

我正在寻找以下查询的答案：客户想要一间连续三个晚上的单间。查找2016年12月的第一个可用日期。

根据问题，这应该是正确的答案。但是我不知道该怎么解决。

+-----+------------+
| id  | MIN(i)     |
+-----+------------+
| 201 | 2016-12-11 |
+-----+------------+

链接来自question number 14 here。

这是数据库的ER图： ER diagram of the database

Answer 1

很抱歉，我对这种查询有点不满意，不能保证所有语法都正确，但是我认为类似以下内容的方法可能会起作用：

SELECT id, DATE_ADD(b.booking_date, INTERVAL (end_date + 1 DAY) as date
FROM  (
         SELECT r.id, STR_TO_DATE('2016-01-01', '%Y-%m-%d') as start_of_month, b.booking_date as start_date, DATE_ADD(b.booking_date, INTERVAL (nights - 1) DAY) as end_date
         FROM room r
         LEFT JOIN booking b ON r.id = b.room_no
         ORDER BY r.id, b.booking_date
      ) as room_bookings
WHERE DATE_DIFF(room_bookings.start_of_month, room_bookings.start_date) >= 3
OR DATE_DIFF(room_bookings.end_date, (
         SELECT b2.booking_date FROM booking b2
         WHERE b2.room_no = room_bookings.id AND b2.booking_date > room_bookings.start_date
         ORDER BY b2.booking_date LIMIT 1)
      ) >= 3

事实上，既然我全都输入了，您也许可以调整主查询的位置，从而甚至不需要room_bookings子选择。希望这会有所帮助，并且不会太远。

Answer 2

这似乎很难在没有日历表的情况下完成-因为一个合适的房间在一个月内可能根本没有预订。没有任何预订，该月开始没有记录。

select r.id, dte
from rooms r cross join
     (select date('2018-12-01') as dte union all
      select date('2018-12-02') as dte union all
      . . .
      select date('2018-12-32') as dte 
     ) d
where not exists (select 1 from bookings b where b.room_no = r.id and b.booking_date = d.dte) and
      not exists (select 1 from bookings b where b.room_no = r.id and b.booking_date = d.dte + interval 1 day) and
      not exists (select 1 from bookings b where b.room_no = r.id and b.booking_date = d.dte + interval 2 day)
order by d.dte
limit 1;

这假设booking_date是中止的开始。您需要提供“单人间”的逻辑。

Answer 3

select distinct top 1 alll.i,alll.room_no,
case
when (select count(*) from booking where room_no = alll.room_no and booking_date between dateadd(day,1,alll.i) and dateadd(day,3,alll.i)) > 0  then 'Y'
else 'N'
end as av3
 from
(select c.i,b.room_no,b.booking_date
from calendar c cross join booking b
where month(c.i) = 12 and year(c.i) = 2016 and b.room_type_requested = 'single'
) as alll
join
(
select distinct c.i, b.room_no
from calendar c join booking b
on c.i between b.booking_date and DATEADD(day,b.nights-1,b.booking_date)
where month(c.i) = 12 and year(c.i) = 2016 and b.room_type_requested = 'single'
) as booked
on alll.i = booked.i
and alll.room_no <> booked.room_no
order by 1

这有效。这有点复杂，但是基本上首先检查所有已预订的房间，然后在一个月的每一天直到接下来的3天之间对未预订的房间进行比较。

Answer 4

我的解决方案是将问题分为两部分（最后是将两个查询连接在一起）。可能不是最有效的，但是解决方案是正确的。

1）在单人间中，查看最后一个退房日期，然后看看哪个空着第一（即该月剩余时间不再预订） 2）检查当前预订之间的间隔-看看它们之间是否有3天的间隔 3）一起加入-抢夺

WITH subquery AS( -- existing single-bed bookings in Dec 
SELECT room_no, booking_date, 
        DATE_ADD(booking_date, INTERVAL (nights-1) DAY) AS last_night
   FROM booking
   WHERE room_type_requested='single' AND
         DATE_ADD(booking_date, INTERVAL (nights-1) DAY)>='2016-12-1' AND 
         booking_date <='2016-12-31'
   ORDER BY room_no, last_night) 

SELECT room_no, MIN(first_avail) AS first_avail --3) join the 2 together
FROM( 

-- 1) check the last date the room is booked in December (available after)

SELECT room_no, MIN(first_avail) AS first_avail
  FROM(
      SELECT room_no, DATE_ADD(MAX(last_night), INTERVAL 1 DAY) AS first_avail
        FROM subquery q3
        GROUP BY 1
        ORDER BY 2) AS t2

UNION 

-- 2) check if any 3-day exist in between reservations 

SELECT room_no, DATE_ADD(MIN(end2), INTERVAL 1 DAY) AS first_avail
  FROM(
      SELECT q1.booking_date AS beg1, q1.room_no, q1.last_night AS end1, 
             q2.booking_date AS beg2, q2.last_night AS end2
        FROM subquery q1
        JOIN subquery q2 
          ON q1.room_no = q2.room_no AND q2.booking_date > q1.last_night
      GROUP BY 2,1
      ORDER BY 2,1) AS t
      WHERE beg2-end1 > 3) AS inner_t

Answer 5

从概念上讲，这是因为第一个可用日期应始终为上次预订的结束。

SELECT MIN(DATE_ADD(a.booking_date, INTERVAL nights DAY)) AS i
FROM booking AS a
WHERE DATE_ADD(a.booking_date, INTERVAL nights DAY)
      >= '2016-12-01'
      AND room_type_requested = 'single'
      AND NOT EXISTS
      (SELECT 1 FROM booking AS b
        WHERE b.booking_date BETWEEN
        DATE_ADD(a.booking_date, INTERVAL nights DAY) 
        AND DATE_ADD(a.booking_date, INTERVAL nights+2 DAY)
        AND a.room_no = b.room_no)

如何查询酒店数据库以返回对连续三个晚上可用的单个房间的查询？

5 个答案: