MySQL - select 按 ID 排列的第二低值

Question

我有一个相当大的数据集，称为offer ，包含大约 7m 行。

该表有 30 列，但我只使用其中的两列， cap_id - 车辆的唯一标识符和price - 租赁车辆的每月成本。

我想编写一个查询，返回每个cap_id的最佳（最低）和次佳价格，以及与次佳价格相比的最佳价格节省百分比。

我正在使用版本 5.7.12

这是SQLFiddle

创建表查询：

CREATE TABLE `offers` (
  `id` mediumint(8) unsigned NOT NULL auto_increment,
  `cap_id` varchar(255) default NULL,
  `price` mediumint default NULL,
  PRIMARY KEY (`id`)
) AUTO_INCREMENT=1;

INSERT INTO `offers` (`cap_id`,`price`) VALUES 
(18452,1007),(18452,884),(18452,276),(90019,328),(73353,539),(64854,249),(26684,257),(37452,966),(90019,980),(73353,1241),
(73353,1056),(37452,1043),(26684,829),(37452,260),(64854,358),(26684,288),(26684,678),(26684,905),(37452,1140),(94826,901),
(90019,745),(37452,1156),(37452,191),(64854,324),(73353,1110),(87725,624),(87725,973),(90019,1203),(90019,709),(18452,1133),
(18452,1019),(37452,639),(37452,1021),(87725,485),(94826,964),(37452,1066),(94826,823),(73353,1056),(18452,621),(37452,272),
(90019,223),(26684,412),(87725,310),(37452,948),(37452,826),(18452,1078),(90019,737),(18452,1166),(73353,150),(73353,1115),
(94826,957),(87725,242),(94826,715),(73353,1190),(94826,320),(94826,869),(64854,574),(94826,505),(26684,322),(90019,949),
(64854,1188),(37452,368),(90019,796),(87725,514),(37452,146),(94826,1216),(18452,625),(64854,1165),(18452,712),(37452,947),
(64854,616),(73353,1065),(26684,1167),(18452,935),(87725,1192),(26684,519),(64854,939),(90019,367),(26684,145),(64854,1076),
(26684,1016),(90019,606),(37452,1066),(73353,609),(94826,343),(94826,236),(94826,1059),(26684,681),(37452,779),(94826,259),
(87725,1080),(37452,914),(90019,826),(37452,597),(26684,879),(87725,471),(94826,680),(18452,906),(87725,860),(94826,1009);

这是我迄今为止尝试过的：

SELECT 
 o1.cap_id,
 o2.price AS best_price,
 o1.price AS next_best,
 (o1.price / o2.price) * 100 AS '%_diff'
FROM
 offers o1
     JOIN
 offers o2 ON o1.cap_id = o2.cap_id
     AND o1.price > o2.price
GROUP BY o1.cap_id
HAVING COUNT(o1.price) = 2

这将返回 0 行，并且当我在我们的数据库中运行它时运行速度非常慢。

这是解释的 output：

ID	选择类型	桌子	分区	类型	可能的键	钥匙	key_len	参考	行	过滤	额外的	现场13	FIELD14
1	简单的	X		指数	cap_id	idx_profile_grouping	idx_capId_monthlyPayment	idx_capId_monthlyPayment	9		7220930	100.00	使用索引； 使用临时的； 使用文件排序
1	简单的	是的		参考	cap_id	idx_profile_grouping	idx_capId_monthlyPayment	idx_capId_monthlyPayment	4	moneyshake.x.cap_id	871	33.33	使用哪里； 使用索引

在此先感谢您的任何建议。

Answer 1

这个过程可以在 MySQL 的较新版本中进行优化，但是由于您的小提琴在 5.6 中，所以我的答案是：

SELECT x.*
     , COUNT(*) running 
  FROM offers x 
  JOIN offers y 
    ON y.cap_id = x.cap_id 
   AND y.price < x.price 
 GROUP 
    BY x.id
 ORDER
    BY x.cap_id 
     , x.price;

扩展这个想法：

SELECT a.*
     , b.price
     , 1-(a.price/b.price) saving
  FROM 
     ( SELECT cap_id
            , MIN(price) price
         FROM offers 
        GROUP  
           BY cap_id
     ) a -- lowest price per cap_id
  JOIN 
     ( SELECT x.cap_id
            , x.price
         FROM offers x
         JOIN offers y
           ON y.cap_id = x.cap_id 
          AND y.price < x.price 
        GROUP 
           BY x.id
       HAVING COUNT(*)= 2
    ) b -- 2nd lowest price per cap_id (other methods are available)
   ON b.cap_id = a.cap_id
ORDER 
   BY a.cap_id;

Answer 2

在 MySQL 8.x 中，您可以执行以下操作：

with
p as (
  select
    id, cap_id, price,
    row_number() over(partition by cap_id order by price) as rn
  from offers
)
select 
  a.id as lowest_id, a.cap_id as lowest_cap_id, a.price as lowest_price,
  b.id as second_id, b.cap_id as second_cap_id, b.price as second_price,
  case when b.price is not null then 
    (b.price - a.price) / b.price
  end as percentage_saving
from p a
left join p b on a.cap_id = b.cap_id and b.rn = 2
where a.rn = 1