每个城市十个最便宜的房间

如何解决每个城市十个最便宜的房间

| 我有3张桌子的房间，酒店，城市，我该如何选择每个城市的十个最便宜的房间及其hotel_id 可以为每个城市执行子查询？

SELECT price,room_id,hotel_id,city_id
    FROM Hotel 
    JOIN  Room USING(hotel_id)
ORDER BY price
LIMIT 10


Scheme
CREATE TABLE `City` (
  `city_id` int(11) NOT NULL AUTO_INCREMENT,`city_name` varchar(50) NOT NULL,PRIMARY KEY (`city_id`)
) 

CREATE TABLE `Hotel` (
  `hotel_id` int(11) NOT NULL AUTO_INCREMENT,`city_id` int(11) NOT NULL,`hotel_name` varchar(100) NOT NULL,PRIMARY KEY (`hotel_id`),CONSTRAINT `Hotel_FK_1` FOREIGN KEY (`city_id`) REFERENCES `City` (`city_id`) ON DELETE CASCADE ON UPDATE CASCADE
)

CREATE TABLE `Room` (
  `room_id` int(11) NOT NULL AUTO_INCREMENT,`hotel_id` int(11) NOT NULL,`room_name` varchar(255) DEFAULT NULL,`room_price1` int(11) DEFAULT NULL,PRIMARY KEY (`room_id`),CONSTRAINT `Room_FK_1` FOREIGN KEY (`hotel_id`) REFERENCES `Hotel` (`hotel_id`) ON DELETE CASCADE ON UPDATE CASCADE
)

用子查询查询

SELECT 
     r.room_price1,r.room_id,h.hotel_name,c.city_name
FROM Hotel h 
INNER JOIN Room r ON
    h.hotel_id = r.hotel_id
INNER JOIN City c USING(city_id)
WHERE (
   SELECT count(*) from Room as rt
   JOIN Hotel ht using(hotel_id)
   where c.city_id = ht.city_id and rt.room_price1 < r.room_price1 OR (rt.room_price1 = r.room_price1 AND rt.room_id < r.room_id)
) < 3 
ORDER BY c.city_name,h.hotel_id,r.room_price1

如果某些酒店有一些room_price1，则此查询返回多于3行的城市信息，并且我需要每个城市3个唯一的酒店，此查询可以从一个酒店返回3个房间 请尝试使用用户变量

set @num := 0,@type := \'\';
select *
from (
   select r.room_price1 pr,c.city_name,@num := if(@type = city_id,@num + 1,1) as row_number,@type := city_id as dummy
  from Hotel h
  JOIN Room r USING(hotel_id)
  INNER JOIN City c USING(city_id)
  order by city_name,r.room_price1 asc
) as x where x.row_number <=3;

对于row_number <6（我的初始化数据），此方法返回错误的排序     

解决方法

那里有很多关于该主题的博客文章。 对我来说最有价值的可能是 Xaprb帖子（有关详细说明，请参见“从每个组中选择前N行”一节） http://www.xaprb.com/blog/2006/12/07/how-to-select-the-firstleastmax-row-per-group-in-sql/ 在您的情况下，这意味着。根据您的需求对该SQL进行建模。

select type,variety,price
from fruits
where (
   select count(*) from fruits as f
   where f.type = fruits.type and f.price < fruits.price
) <= 2;

让我们尝试一下

SELECT 
   r.price,r.room_id,h.hotel_id,h.city_id
FROM Hotel h INNER JOIN Room r ON
  h.hotel_id = r.hotel_id
WHERE (
   select count(*) from Room as rt
   where r.hotel_id = rt.hotel_id and r.price > rt.price
) <= 10;
ORDER BY h.city_id,r.price

如您所见，您可以进行子查询以选择所有房间的数量 价格低于当前根的价格，看看是否 计数小于或等于10。 特别要注意我如何改变

f.price < fruits.price

至

r.price > rt.price

因为在Xaprb的示例中，他选择了N个最便宜的水果 希望能有所帮助。 BTW Stackoverflow并非用于解决作业。您也应该学习一些自己的东西；-)     ,我假设表“ 7”引用了它所属的“ 8”

SELECT price,room_id,hotel_id,city_id
    FROM Hotel,Room 
    WHERE Hotel.hotel_id = Room.hotel_id
ORDER BY price GROUP BY city_id
LIMIT 10