查询以获取上周销售额最高的产品获得者

Question

我有一个包含三列的数据库 table。

WeekNumber、ProductName、SalesCount

示例数据如下所示table。我想要第 26 周比前一周（即第 25 周）排名前 10 的获利者（按百分比）。唯一的条件是产品在这两个星期的销售额都应大于 0。

示例数据中B、C、D为常见产品，C的增益百分比最高。

同样，我也需要前 10 名失败者。

到目前为止我尝试的是在两周内进行内部连接并获得共同产品。但是，我无法获得最高收益者的逻辑。

输出应该像

 Product    PercentGain

  C            400%

  D            12.5%

  B            10%

Answer 1

这会给你一个通用的答案，而不仅仅是针对任何特定的一周：

select top 10 product , gain [gain%]
from 
(
SELECT product, ((curr.salescount-prev.salescount)/prev.salescount)*100 gain 
from   
  (select weeknumber, product, salescount from tbl) prev
  JOIN
  (select weeknumber, product, salescount from tbl) curr
on prev.weeknumber = curr.weeknumber - 1
AND prev.product = curr.product
where prev.salescount > 0 and curr.salescount > 0
)A 
order by gain desc

如果您对第 25 周和第 26 周感兴趣，则只需在 WHERE 子句中添加以下条件：

and prev.weeknumber = 25

Answer 2

如果您使用的是 SQL-Server 2012（或更新版本），您可以使用 lag 函数将 "this" 周的销售额与前一周的销售额相匹配。从那以后，这只是一些数学：

SELECT   TOP 10 product, sales/prev_sales - 1 AS gain
FROM     (SELECT product, 
                 sales, 
                 LAG(sales) OVER (PARTITION BY product 
                                  ORDER BY weeknumber) AS prev_sales
          FROM   mytable) t
WHERE    weeknumber = 26 AND
         sales > 0 AND
         prev_sales > 0 AND
         sales > prev_sales
ORDER BY sales/prev_sales

Answer 3

这是查询。

select top 10 product , gain [gain%]
from 
(
SELECT curr.Product, ( (curr.Sales - prev.Sales ) *100)/prev.Sales gain 
from   
  (select weeknumber, product, sales from ProductInfo where weeknumber = 25 ) prev
  JOIN
  (select weeknumber, product, sales from ProductInfo where weeknumber = 26 ) curr
on    prev.product = curr.product
where prev.Sales > 0 and curr.Sales > 0
)A 
order by gain desc

查询以获取上周销售额最高的产品获得者

Query to get top product gainers by sales over previous week

sql

sql-server

select

top-n