根据具有条件的组对连续月份求和

Question

我无法缩小连续几个月发生的主要地区的销售额。我知道我需要将某种形式的 window 函数与 Row_Number 或 Dense_Rank 一起使用，但我无法获得最终输出

这是我的源数据：

+--------+-----------+------------+
| Fruit  | SaleDate  | Top_Region |
+--------+-----------+------------+
| Apple  | 1/1/2017  |          1 |
| Apple  | 2/1/2017  |          1 |
| Apple  | 3/1/2017  |          1 |
| Apple  | 4/1/2017  |          0 |
| Apple  | 5/1/2017  |          0 |
| Apple  | 6/1/2017  |          0 |
| Apple  | 7/1/2017  |          1 |
| Apple  | 8/1/2017  |          1 |
| Apple  | 9/1/2017  |          1 |
| Apple  | 10/1/2017 |          1 |
| Apple  | 11/1/2017 |          0 |
| Apple  | 12/1/2017 |          0 |
| Banana | 1/1/2017  |          0 |
| Banana | 2/1/2017  |          0 |
| Banana | 3/1/2017  |          1 |
| Banana | 4/1/2017  |          1 |
| Banana | 5/1/2017  |          1 |
| Banana | 6/1/2017  |          1 |
| Banana | 7/1/2017  |          1 |
| Banana | 8/1/2017  |          1 |
| Banana | 9/1/2017  |          0 |
| Banana | 10/1/2017 |          1 |
| Banana | 11/1/2017 |          1 |
| Banana | 12/1/2017 |          0 |
+--------+-----------+------------+

这是预期的输出：

+--------+-----------+-----------+-------+
| Fruit  |   Start   |    End    | Total |
+--------+-----------+-----------+-------+
| Apple  | 1/1/2017  | 3/1/2017  |     3 |
| Apple  | 7/1/2017  | 10/1/2017 |     4 |
| Banana | 3/1/2017  | 8/1/2017  |     6 |
| Banana | 10/1/2017 | 11/1/2017 |     2 |
+--------+-----------+-----------+-------+

目标是在一个月内连续出现销量第一的区域，但没有出现。

到目前为止，我已经尝试了几种不同的组合，这是最接近的。

SELECT fruit,
        MIN(saledate) AS spanStart ,
        MAX(saledate) AS spanEnd,
        COUNT(*) AS spanLength
FROM    ( SELECT    s.* ,
                    ( ROW_NUMBER() OVER ( ORDER BY month )
                      - ROW_NUMBER() OVER ( PARTITION BY fruit, topregion ORDER BY month ) ) AS fruits
          FROM      #salesdata s
        ) s
GROUP BY fruit,fruits ,
        topregion
HAVING  topregion = 1
ORDER BY COUNT(*) DESC;

如有任何帮助，我们将不胜感激

Answer 1

这是一个典型的缺口孤岛问题。一种策略是通过计算两个 row_number() 之间的差异来识别相邻行组的组。然后我们可以过滤具有 top_region = 1 的组并使用聚合来获取开始日期、结束日期和每个组的记录数。

您的查询非常接近，但第一个 row_number() 在其 over() 子句中缺少 partition by fruit。而且我发现给另一列称为 fruit 的列 fruits 取别名很容易出错。

select 
    fruit,
    min(sale_date) start_date,
    max(sale_date) end_date,
    count(*) total
from (
    select 
        t.*,
        row_number() over(partition by fruit order by sale_date) rn1,
        row_number() over(partition by fruit, top_region order by sale_date) rn2
    from mytable t
) t
where top_region = 1
group by fruit, rn1 - rn2
order by fruit, start_date

您可以运行单独的内部查询来查看它产生的结果。

Demo on DB Fiddle:

fruit  | start_date | end_date   | total
:----- | :--------- | :--------- | ----:
Apple  | 2017-01-01 | 2017-01-03 |     3
Apple  | 2017-01-07 | 2017-01-10 |     4
Banana | 2017-01-03 | 2017-01-08 |     6
Banana | 2017-01-10 | 2017-01-11 |     2

根据具有条件的组对连续月份求和

Sum Consecutive Months Based on Groups with Criteria

sql-server

aggregate-functions

window-functions

gaps-and-islands

ranking-functions