Question

我正在尝试将其整合到一个查询中。

预订系统有2张桌子，酒店和预订。

hotels(id, num_rooms)
reservations(hotel_id, start_date, end_date) (each reservation is for 1 room)

我需要执行一个查询，其中用户说他们在date1和date2之间需要x个房间，并且系统返回所有具有足够房间的酒店（每天计算所有预订以确定可用房间）的日期。

我的实验一团糟，因为我无法想到如何将这些子结果组合在一起以取得一些进展。我知道我将不得不使用一些子查询来实现这一目标，我无法正确地绕过它。

   -- a poor snippet to try and get some foothold
   SELECT h.* from hotels as h 
   left join reservations as r on r.hotel_id=h.id 
   where 
   --(select count(*) from reservations where start date between '2015-01-01' and '2015-01-31' group by hotel_id)

Answer 1

此解决方案生成所有预订的日期范围，并计算每个酒店和日期的预订数量。然后，查询使用此信息检查房间总数减去请求期间内任何一天的最大预订数等于或高于所请求的房间数。

我使用派生表来帮助生成日期范围，但可以很容易地用数字表替换。

我的脚本（我使用变量作为查询限制）：

set @start_date := '2015-04-01'; 
set @end_date   := '2015-04-04'; 
set @num_rooms  := 1;

SELECT 
    id AS hotel_id
FROM
    hotels h
        LEFT JOIN
    (SELECT 
        hotel_id, date, COUNT(*) reserved
    FROM
        (SELECT 
        r.hotel_id, a.date, r.start_date, r.end_date
    FROM
        (SELECT 
        CAST(@start_date + INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY
                AS DATETIME) AS Date
    FROM
        (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a
    CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b
    CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c) a
    INNER JOIN reservations r ON a.Date >= @start_date
        AND a.Date <= @end_date
    LEFT JOIN hotels h ON h.id = r.hotel_id) reservations
    WHERE
        date >= start_date AND date <= end_date
    GROUP BY hotel_id , date) a ON a.hotel_id = h.id
GROUP BY hotel_id , num_rooms
HAVING (num_rooms - MAX(reserved)) >= @num_rooms
    OR hotel_id IS NULL

我的测试数据：

create table hotels (id int, num_rooms int);
insert hotels values (1, 3),(2, 5), (3, 20);
create table reservations (hotel_id int, start_date date, end_date date);
insert reservations (hotel_id, start_date, end_date) values
(1, '2015-03-01', '2015-03-05'),
(1, '2015-03-01', '2015-03-02'),
(1, '2015-03-01', '2015-03-05'),

(2, '2015-03-01', '2015-03-05'),
(2, '2015-03-01', '2015-03-05'),
(2, '2015-03-01', '2015-03-05'),
(2, '2015-03-02', '2015-03-05'),

(3, '2015-03-01', '2015-03-05'),
(3, '2015-03-02', '2015-03-05'),
(3, '2015-03-03', '2015-03-05'),
(3, '2015-03-04', '2015-03-06');

使用上面的测试数据，查询将返回酒店2和3。

MYSQL预订搜索，查询两个日期之间的每一天

1 个答案: