试图获得在查询中获利最多的酒店的房间

问题描述

我正在尝试执行一个查询，该查询为我提供了每家酒店(25家酒店)中最赚钱的客房的总和

I'm trying to implement a query which give me the sum of most profitable room in each hotel(25 hotels)

以下是我的查询:

SELECT hotels.hotel_id,rooms.room_id,hotel_name,room_number,sum(rooms.room_price) AS profit,COUNT(rooms.room_id) AS countFROM hotels,rooms,bookingsWHERE hotels.hotel_id=rooms.hotel_idAND rooms.room_id=bookings.room_idGROUP BY rooms.room_id

SELECT hotels.hotel_id,rooms.room_id,hotel_name,room_number,sum(rooms.room_price) AS profit,COUNT(rooms.room_id) AS countFROM hotels,rooms,bookingsWHERE hotels.hotel_id=rooms.hotel_idAND rooms.room_id=bookings.room_idGROUP BY rooms.room_id

这是我得到的最接近的结果..忽略酒店名称语言

and this is the closest outcome i got.. ignore the hotel name language

这是我到目前为止所达到的结果，

酒店

房间

预订第1页

预订第2页(剩余记录)

hotel_id 1有5个房间，其中room_number 300利润最大.我只想显示每家酒店的最大利润.我不需要其他利润较少的房间.

hotel_id 1 has 5 rooms, the room_number 300 made the most profit. I want to show the most profit only of each hotel. I don't need the other rooms that made less profit.

更新:因此，我解决了一个类似的查询，在该查询中，我想获得获利最多的最好的2个房间.但是，我只是想不出任何能给我每家酒店带来最大利益的功能.很少的提示或帮助就会被应用

Update:So i solved a similar query where i want to get the best 2 rooms that made the most profit. But, i just can't think of any function that can give me only best profit of each hotel. little hint or help would be appriciated

推荐答案

尝试以下查询:

SELECT * FROM 
(SELECT hotels.hotel_id,rooms.room_id,hotel_name,room_number,SUM(rooms.room_price) AS profit,COUNT(rooms.room_id) AS COUNT
FROM hotels,rooms,bookings
WHERE hotels.hotel_id=rooms.hotel_id
AND rooms.room_id=bookings.room_id
GROUP BY rooms.room_id) a GROUP BY hotel_id;

这可能会做到:

SELECT hotel_id,room_id,room_number,MAX(a.tc) AS "Count",MAX(tp) AS "MostProfit" FROM
(SELECT hotel_id,rooms.room_id,room_number,COUNT(rooms.room_id) AS "tc",SUM(room_price) AS "tp" FROM rooms JOIN bookings 
ON rooms.room_id=bookings.room_id
GROUP BY rooms.room_id) a GROUP BY hotel_id

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