质疑上周按销售额计算的顶级产品涨幅

Question

我有一个包含三列的数据库表。

WeekNumber，ProductName，SalesCount

样本数据如下表所示。我希望在第26周比前一周（即第25周）获得前10名获利者（按％计算）。唯一的条件是该产品在两周内的销售数量应大于0。

在样本数据B中，C，D是常见产品，C具有最高的％增益。

同样，我也需要十大输家。

我到目前为止尝试的是在两周之内进行内部联接并获得常见产品。但是，我无法获得顶级赢家的逻辑。

输出应该像

 Product    PercentGain

  C            400%

  D            12.5%

  B            10%

Answer 1

这将为您提供一般答案，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，那么只需在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），则可以使用lag功能将“此”周销售额与前一周相匹配。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