结合两个查询sql答案

【问题标题】：Combining two queries sql结合两个查询sql
【发布时间】：2018-04-09 08:45:59
【问题描述】：

我不知道如何组合以下查询

查询 1

SELECT hotel.country, time.year, time.month, COUNT(booking.room_id) as booked 

FROM booking 
LEFT JOIN room on room.room_id = booking.room_id 

LEFT JOIN hotel on room.hotel_id = hotel.hotel_id 

LEFT JOIN time on booking.time_id = time.time_id 

GROUP BY hotel.country, time.year, time.month 

ORDER by hotel.country, time.year, time.month

查询 2

SELECT hotel.country, time.year, time.month, COUNT(checkout.room_id) as checkedout

FROM checkout

LEFT JOIN room on room.room_id = checkout.room_id

LEFT JOIN hotel on room.hotel_id = hotel.hotel_id

LEFT JOIN time on checkout.time_id = time.time_id

GROUP BY hotel.country, time.year, time.month

ORDER BY hotel.country, time.year, time.month

我希望输出如下

国家、年份、月份、预订次数、结账次数、总和（预订+结账）

有什么建议吗？

【问题讨论】：

联合所有。聚合。
为什么是外连接？在 room_id 或 hotel_id 或 time_id 为空的情况下是否存在预订和结账？这似乎没有多大意义。
找工会？？？
由于booking 指的是某个time，我推测这是预订的时间，而不是预订的时间（因为那将是一个时间范围代替）。正确的？因此，我们可能会找到有预订但没有结账的月份，以及有结账但没有预订的月份。是吗？

标签： mysql sql

【解决方案1】：

使用 mysql Union 和聚合。

联合文档：https://dev.mysql.com/doc/refman/5.7/en/union.html 汇总文档：https://dev.mysql.com/doc/refman/5.7/en/group-by-functions.html

下面的联合示例：

SELECT 
    firstName, 
    lastName
FROM
    employees 
UNION 
SELECT 
    contactFirstName, 
    contactLastName
FROM
    customers;

希望这会有所帮助！

【讨论】：

【解决方案2】：

SELECT T1.country, 
       T1.year, 
       T1.month,
       T1.booked,
       T2.checkedout, 
      (T1.booked + T2.checkedout) AS 'sum(booking+checkout)'
FROM 
(
SELECT hotel.country, time.year, time.month, COUNT(booking.room_id) as booked 

FROM booking 
LEFT JOIN room on room.room_id = booking.room_id 

LEFT JOIN hotel on room.hotel_id = hotel.hotel_id 

LEFT JOIN time on booking.time_id = time.time_id 

GROUP BY hotel.country, time.year, time.month 

ORDER by hotel.country, time.year, time.month
) AS T1
INNER JOIN
(
SELECT hotel.country, time.year, time.month, COUNT(checkout.room_id) as checkedout

FROM checkout

LEFT JOIN room on room.room_id = checkout.room_id

LEFT JOIN hotel on room.hotel_id = hotel.hotel_id

LEFT JOIN time on checkout.time_id = time.time_id

GROUP BY hotel.country, time.year, time.month

ORDER BY hotel.country, time.year, time.month
)AS T2
ON T1.country=T2.country
AND T1.year=T2.year
AND T1.month=T2.month

【讨论】：

【解决方案3】：

解决此问题的一种方法是使用UNON ALL：

SELECT h.country, t.year, t.month, SUM(is_booking) as booked, SUM(is_checkout) as checkedout
FROM ((SELECT b.room_id, b.time_id, 1 as is_booking 0 as is_checkout
       FROM booking b
      ) UNION ALL
      (SELECT c.room_id, c.time_id, 0 as is_booking, 1 as is_checkout
       FROM checkout c
      )
     ) bc  LEFT JOIN
     room r
     ON r.room_id = bc.room_id LEFT JOIN
     hotel h
     ON r.hotel_id = h.hotel_id LEFT JOIN
     time t
     ON bc.time_id = t.time_id 
GROUP BY h.country, t.year, t.month 
ORDER by h.country, t.year, t.month

【讨论】：