SQL - 连接 2 个表并返回 2 行之间的值差异

Posted 2023-02-25

【中文标题】SQL - 连接 2 个表并返回 2 行之间的值差异

【英文标题】：SQL - Join 2 tables and also return difference in value between 2 rows

【发布时间】：2021-11-05 00:21:55

【问题描述】：

我正在尝试返回具有各种条件（搜索词、日期等）的产品列表，并且我有以下示例 SQL。

SELECT * 
FROM products  
JOIN ( SELECT date, price, product_id 
       FROM pricing 
       WHERE id IN ( (SELECT  MAX(id) 
                      FROM pricing 
                      GROUP BY product_id )  )  
        AND date = '2021-09-08' ) as current_price on products.id=price.product_id 
WHERE 1 
  AND sku like '%%'   
   OR title like '%%'  
ORDER BY id ASC

它从 2 个表中提取数据：产品和定价。

产品

+----+----------------+--------+
| id |     title      |  sku   |
+----+----------------+--------+
|  1 | Product Name 1 | 654987 |
|  2 | Product Name 2 | 548879 |
|  3 | Product Name 3 | 541685 |
+----+----------------+--------+

定价

+----+------------+------------+-------+
| id | product_id |    date    | price |
+----+------------+------------+-------+
|  1 |          1 | 06-09-2021 | 1.99  |
|  2 |          2 | 06-09-2021 | 1.99  |
|  3 |          1 | 07-09-2021 | 3.99  |
|  4 |          3 | 06-09-2021 | 5.99  |
|  5 |          3 | 08-09-2021 | 6.99  |
|  6 |          1 | 08-09-2021 | 6.99  |
+----+------------+------------+-------+

期望的输出

+----+----------------+--------+---------------+-------------------+------------+
| id |     title      |  sku   | current_price | last_price_update | difference |
+----+----------------+--------+---------------+-------------------+------------+
|  1 | Product Name 1 | 654987 | 6.99          | 08-09-2021        | 2.5        |
|  2 | Product Name 2 | 548879 | 1.99          | 06-09-2021        | 0         |
|  3 | Product Name 3 | 541685 | 6.99          | 08-09-2021        | -1          |
+----+----------------+--------+---------------+-------------------+------------+

我的问题是我也想返回一个差异列，显示最新价格与之前的价格之间的差异。

我已经尝试了十亿次，但 SQL 不是我的菜，我要么返回完整的价格列表，要么什么都不返回。

使用 MariaDB 版本：10.3.30

谢谢， 阿什利

【问题讨论】：

精确的 mysql 版本是...？以 CREATE TABLE + INSERT INTO（或创建在线小提琴）的形式提供示例数据，并提供此数据的所需输出并附上解释。

【参考方案1】：

您可以使用row_number() 枚举价格，然后条件聚合得到两个价格：

select p.*, pr.current_price, pr.previous_price,
       (pr.current_price - pr.previous_price) as diff
from products p join
     (select product_id,
             max(case when seqnum = 1 then price end) as current_price,
             max(case when seqnum = 2 then price end) as previous_price
      from (select pr.*,
                   row_number() over (partition by product_id order by date desc) as seqnum
            from pricing pr
           ) pr
      where seqnum <= 2
      group by product_id
      ) pr
     on pr.product_id = p.id;