我希望能够比较两个不同日期的产品收入答案

【问题标题】：I want to be able to compare revenue for products from two different dates我希望能够比较两个不同日期的产品收入
【发布时间】：2022-11-19 12:43:23
【问题描述】：

所以我可以像这样运行两个单独的查询：

SELECT date as date1, product as product1, product_id as product_id_1, SUM(revenue) AS rev1 
FROM product_inventory 
WHERE date = '2021-11-17' 
GROUP BY date1 , product1, product_id_1 
ORDER BY rev1 DESC

SELECT date as date2, product as product2, product_id as product_id_2, SUM(revenue) AS rev2 
FROM product_inventory 
WHERE date = '2022-11-17' 
GROUP BY date2 , product2, product_id_2 
ORDER BY rev2 DESC

这是我为每个人得到的输出：

date1	product1	product_id_1	rev1
2021-11-17	adidas samba	9724	6087.7000732421875
2021-11-17	nike air max	5361	4918.0
2021-11-17	puma suede	1985	3628.1600341796875

date2	product2	product_id_2	rev2
2022-11-17	adidas samba	9724	5829.0
2022-11-17	nike air max	5361	4841.864013671875
2022-11-17	puma suede	1985	5404.4140625

我怎样才能以一种将 date2 和 rev2 列拉成这样的单个输出的方式查询数据库？

date1	product1	product_id_1	rev1	date2	rev2
2021-11-17	adidas samba	9724	6087.7000732421875	2022-11-17	5829.0
2021-11-17	nike air max	5361	4918.0	2022-11-17	4841.864013671875
2021-11-17	puma suede	1985	3628.1600341796875	2022-11-17	5404.4140625

我试过这个查询：

SELECT A.date1, A.product1, A.rev1, B.date2, B.product2, B.rev2 FROM
(
SELECT date as date1, product as product1, product_id as product_id_1, SUM(revenue) AS rev1 FROM product_inventory WHERE date = '2021-11-17' GROUP BY date1 , product1, product_id_1 ORDER BY rev1 DESC
) A,
(
SELECT date as date2, product as product2, product_id as product_id_2, SUM(revenue) AS rev2 FROM product_inventory WHERE date = '2022-11-17' GROUP BY date2, product2, product_id_2 ORDER BY rev2 DESC
) B;

但我得到了这个输出

date1	product1	rev1	date2	product2	rev2
2021-11-17	puma suede	3628.1600341796875	2022-11-17	adidas samba shoes	5829.0
2021-11-17	nike air max	4918.0	2022-11-17	adidas samba shoes	5829.0
2021-11-17	adidas samba	6087.7000732421875	2022-11-17	adidas samba shoes	5829.0
2021-11-17	puma suede	3628.1600341796875	2022-11-17	puma suede	5404.4140625
2021-11-17	nike air max	4918.0	2022-11-17	puma suede	5404.4140625
2021-11-17	adidas samba	6087.7000732421875	2022-11-17	puma suede	5404.4140625
2021-11-17	puma suede	3628.1600341796875	2022-11-17	nike air max	4841.864013671875
2021-11-17	nike air max	4918.0	2022-11-17	nike air max	4841.864013671875
2021-11-17	adidas samba	6087.7000732421875	2022-11-17	nike air max	4841.864013671875

这就像记录数的平方。

【问题讨论】：

是的，from tablea, tableb 检查 tablea 和 tableb 行的所有可能组合，而不是您想要的。 select version(); 显示什么？

标签： mysql sql database

【解决方案1】：

您需要像这样在 product_id 上使用 JOIN：

WITH DATE_1 AS (
SELECT date as date1, product as product1, product_id as product_id_1, SUM(revenue) AS rev1 
  FROM product_inventory 
 WHERE date = '2021-10-17' 
 GROUP BY 1, 2, 3
),
DATE_2 AS (
SELECT date as date2, product as product2, product_id as product_id_2, SUM(revenue) AS rev2 
  FROM product_inventory 
 WHERE date = '2021-11-17' 
 GROUP BY 1, 2, 3
)
SELECT D1.*, D2.*
  FROM DATE_1 D1
       INNER JOIN DATE_2 D2
          ON D1.product_id_1 = D2.product_id_2

【讨论】：

【解决方案2】：

您正在获得表 A 和 B 的笛卡尔连接，即所有行的组合来自一个和乙，因此您会看到记录增加到 9 行，因为 3 x 3。

你需要做的是做一个表的连接一个和乙关于产品：例如左连接：

SELECT A.*, B.*
FROM A
LEFT JOIN B on A.product = B.product

尽可能避免在 SELECT 中使用 *。

【讨论】：

【解决方案3】：

你需要加入两个结果集。像这样：

SELECT t1.date, t1.product, t1.product_id, t2.date, t2.product_id, ...
FROM
  (SELECT ... FROM product_inventory ...) AS t1
JOIN
  (SELECT ... FROM product_inventory ...) AS t2
ON t1.product_id = t2.product_id AND t1.date = t2.date

此外，MySQL 8 引入了一个名为 Common Table Expressions (CTE) 的功能，它可能是上述查询的一个很好的替代方案，因为它允许您将要连接的两个结果集的 SELECT 查询分开。所以使用 CTE 你可以这样写：

WITH 
  t1 AS
  (SELECT ... FROM product_inventory ...),
  t2 AS
  (SELECT ... FROM product_inventory ...),
SELECT t1.date, t1.product, t1.product_id, t2.date, t2.product_id, ...
FROM t1      
JOIN t2 ON t1.product_id = t2.product_id AND t1.date = t2.date

【讨论】：