过去 N 周的 SQL 销售额总和，包括销售额为 0 的一周

Question

我想显示每件商品每周 4 周的销售额总和（当前周加上前 3 周）。

原始数据

+--------+--------+----------+
| Item   | Week   | sales    |
+--------+--------+----------+
|   a    |      1 |     10   |
|   a    |      2 |     10   |
|   a    |      4 |     10   |
|   a    |      7 |     10   |
|   a    |      8 |     10   |
|   a    |     10 |     10   |
|   b    |      1 |     10   |
|   b    |      2 |     10   |
|   b    |      4 |     10   |
|   b    |      7 |     10   |
|   b    |      8 |     10   |
|   b    |     10 |     10   |
+--------+--------+----------+

预期结果（以a项为例）

+------+------+------------------------------------------------------+
| Item | Week |                        sales                         |
+------+------+------------------------------------------------------+
| a    |    1 | 10                                                   |
| a    |    2 | 20                                                   |
| a    |    3 | 30                                                   |
| a    |    4 | 30                                                   |
| a    |    5 | 20(Note: sales of Week 5+Week 4+Week 3+Week 2)       |
| a    |    6 | 10                                                   |
| a    |    7 | 20[Note: 10(Week 7)+0(Week 6)+0(Week 5)+10 (Week 4)] |
| a    |    8 | 20                                                   |
| a    |    9 | 20                                                   |
| a    |   10 | 30(Note: sales of Week 10+Week 9+Week 8+Week 7)      |
+------+------+------------------------------------------------------+

我尝试使用

sum (sales) over (partition by item order by week row 3 preceding)

但是，它会跳过没有销售记录的周，并将所有非零值相加。例如：对于第 7 周，sum over 计算结果为 40，因为它将第 7、4、2、1 周的销售额相加。 有没有办法达到预期的效果？

Answer 1

如果您只需要查看现有周的总和（因此不填充缺失的周），这可以在标准 SQL 中通过使用范围而不是行来实现：

select item, 
       week,
       sum (sales) over (partition by item order by week range between 3 preceding and current row)
from the_table
order by item, week;

如果您需要以行的形式填写缺失的周数，答案很大程度上取决于所使用的 DBMS。在 Postgres 中，您可以执行以下操作：

with items as (
  select distinct item
  from data
), all_weeks as (
  select i.item, g.week
  from items i
    cross join generate_series(1,10) as g(week)
)    
select t.item,
       t.week,
       sum (d.sales) over (partition by t.item order by t.week range between 3 preceding and current row)
from all_weeks t
  left join data d on d.item = t.item and d.week = t.week

对于其他 DBMS 系统，generate_series() 调用可以替换为固定的周数：

select i.item, g.week
from items i
  cross join ( values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10) ) as g(week) 

Online example

Answer 2

由于问题没有提到 RDBMS，我在 SQL Server 中测试了下面的代码。以下解决方案适用于 SQL Server

我正在做以下活动：

1. 生成周、项目的所有组合
2. 对于组合，生成前 4 周的总和（销售额）
3. 删除原始列表中不存在的周

DECLARE @table TABLE (Item CHAR(1), Week TINYINT, sales INT)

INSERT INTO @table
VALUES ('a', 1, 10), ('a', 2, 10), ('a', 4, 10), ('a', 7, 10), ('a', 8, 10), ('a', 10, 10), ('b', 1, 10), ('b', 2, 10), ('b', 4, 10), ('b', 7, 10), ('b', 8, 10), ('b', 10, 10);;

WITH CTE_ItemWeek
AS (
    SELECT Item, Week
    FROM (
        VALUES (1), (2), (3), (4), (5), (6), (7), (8), (9), (10)
        ) AS t(week)
    CROSS JOIN (
        SELECT DISTINCT Item
        FROM @table
        ) A
    )
SELECT *
FROM (
    SELECT t1.Item, CASE 
            WHEN t2.Week IS NULL
                THEN LAG(t2.week) OVER (
                        PARTITION BY t1.Item ORDER BY t2.week
                        )
            ELSE t2.week
            END AS week, SUM(t2.sales) OVER (
            PARTITION BY t1.Item ORDER BY t1.week ROWS BETWEEN 3 preceding
                    AND CURRENT ROW
            ) AS total_sales
    FROM CTE_ItemWeek AS t1
    LEFT OUTER JOIN @table AS t2 ON t1.Item = t2.Item
        AND t1.week = t2.week
    ) AS t
WHERE week IS NOT NULL

结果集

+------+------+-------------+
| Item | week | total_sales |
+------+------+-------------+
| a    |    1 |          10 |
| a    |    2 |          20 |
| a    |    4 |          30 |
| a    |    7 |          20 |
| a    |    8 |          20 |
| a    |   10 |          30 |
| b    |    1 |          10 |
| b    |    2 |          20 |
| b    |    4 |          30 |
| b    |    7 |          20 |
| b    |    8 |          20 |
| b    |   10 |          30 |
+------+------+-------------+

Answer 3

每周，如果你的桌子是桌子

SELECT sum(B.sales), A.week as curweek As totalsales
FROM table A, table B
WHERE curweek - B.week <4 
AND curweek - B.week >=0
GROUP BY curweek

Answer 4

SELECT week,Item ,sum(sales) WHERE (curweek - week < 4) group by week,Item