2016-11-09 46 views
3

我正在使用SQL Server 2012为销售代理提取滚动销售信息。 如果这些代理商在6天或更少的滚动时间内达到15次销售额,则会获得奖金。如果他们成为目标,则滚动计数会重置。 周日应该被忽略。销售目标使用重置的运行总计SQL Server 2012

因此,考虑以下AGENTID,日期和销售数据:

SELECT 1 AgentID,'2016-10-31' Date,1 Sales 
INTO #Sales 
UNION SELECT 1,'2016-11-01',2 
UNION SELECT 1,'2016-11-02',1 
UNION SELECT 1,'2016-11-03',5 
UNION SELECT 1,'2016-11-04',3 
UNION SELECT 1,'2016-11-05',2 
UNION SELECT 1,'2016-11-07',6 
UNION SELECT 1,'2016-11-08',5 
UNION SELECT 1,'2016-11-09',4 
UNION SELECT 1,'2016-11-10',6 
UNION SELECT 1,'2016-11-11',1 
UNION SELECT 1,'2016-11-12',3 
UNION SELECT 1,'2016-11-14',2 
UNION SELECT 1,'2016-11-15',2 
UNION SELECT 1,'2016-11-16',4 
UNION SELECT 1,'2016-11-17',2 
UNION SELECT 1,'2016-11-18',2 

,我期望的目标被击中的日期是:

2016-11-07 (period 2016-11-01 -> 2016-11-07) 
2016-11-10 (period 2016-11-08 -> 2016-11-10) 
2016-11-18 (period 2016-11-12 -> 2016-11-18) 

AgentID Date Sales Qualify 
------------------------------- 
1 2016-10-31 1 0 
1 2016-11-01 2 0 
1 2016-11-02 1 0 
1 2016-11-03 5 0 
1 2016-11-04 3 0 
1 2016-11-05 2 0 
1 2016-11-07 6 1 
1 2016-11-08 5 0 
1 2016-11-09 4 0 
1 2016-11-10 6 1 
1 2016-11-11 1 0 
1 2016-11-12 3 0 
1 2016-11-14 2 0 
1 2016-11-15 2 0 
1 2016-11-16 4 0 
1 2016-11-17 2 0 
1 2016-11-18 2 1 

我已经尝试了几种方法,但我找不到重置滚动总数的方法。

我认为窗口函数是要走的路。

裁减员额像 Window Functions - Running Total with reset

我认为这是类似于我所需要的,但不能完全得到它才能正常工作。

更新: 我尝试的第一件事是创建滚动6天的窗口,但我没有看到这在一个基于集合的方法工作。 我可以使用光标来遍历这些行,但我真的不喜欢这个想法。

SELECT DATEADD(DAY,-6,a.Date) StartDate,Date EndDate,a.AgentID,a.Sales, 
(SELECT SUM(b.Sales) 
    FROM cteSales b 
    WHERE b.Date <= a.Date 
    AND b.Date >= DATEADD(DAY,-6,a.Date)) TotalSales 
FROM cteSales a 

然后我试图用在上面的URL中提到的脚本,但我真的不知道它在做什么。 我只是在改变一些东西,希望在解决方案中陷入困境,而这只是不起作用。

WITH c1 as 
(
    select *, 
    sum(sales) over(order by IDDate rows unbounded preceding) as rt 
    from cteSales 
) 

SELECT date, sales, rt, 
    SalesTarget_rt - lag(SalesTarget_rt, 1, 0) over(order by date) as SalesTarget, 
    rt * SalesTarget_rt as new_rt 

from c1 
    cross apply(values(case when rt >= 15 then 1 else 0 end)) as a1(SalesTarget_rt); 
+0

你有你至今尝试过的脚本? – iamdave

+1

另外,同一天15日之后的销售额是否会计入下一个目标?因此,如果他们在第1天售出14,然后在第2天售出3,那么对第3天的第2个或第2个目标开始计数? – iamdave

+0

哦,还有(!)是否每个代理每天最多只有一条记录? – iamdave

回答

1

那就好!这是一个有趣的挑战,我非常高兴,因为我破解了它。注释等在代码注释中。如果您想更改可累积奖金的天数,请更改@DaysInBonusPeriod中的值。这也适用于多个AgentID秒和日期的任何序列,假设其中的任何缺少的日期不包括奖金计提时期 - 即:如果你忽视周日周三时段进行计数:

Day  Period Day 
Monday 1 
Tuesday 2 
Thursday 3 
Friday 4 
Saturday 5 
Monday 6 

解决方案

declare @t table(AgentID int 
       ,DateValue Date 
       ,Sales int 
       ); 
insert into @t     
select 1,'2016-10-31',1 union all 
select 1,'2016-11-01',2 union all 
select 1,'2016-11-02',1 union all 
select 1,'2016-11-03',5 union all 
select 1,'2016-11-04',3 union all 
select 1,'2016-11-05',2 union all 
select 1,'2016-11-07',6 union all 
select 1,'2016-11-08',5 union all 
select 1,'2016-11-09',4 union all 
select 1,'2016-11-10',6 union all 
select 1,'2016-11-11',1 union all 
select 1,'2016-11-12',3 union all 
select 1,'2016-11-14',2 union all 
select 1,'2016-11-15',2 union all 
select 1,'2016-11-16',4 union all 
select 1,'2016-11-17',2 union all 
select 1,'2016-11-18',2 union all 

select 2,'2016-10-31',1 union all 
select 2,'2016-11-01',7 union all 
select 2,'2016-11-02',0 union all 
select 2,'2016-11-03',0 union all 
select 2,'2016-11-04',0 union all 
select 2,'2016-11-05',0 union all 
select 2,'2016-11-07',0 union all 
select 2,'2016-11-08',0 union all 
select 2,'2016-11-09',1 union all 
select 2,'2016-11-10',3 union all 
select 2,'2016-11-11',2 union all 
select 2,'2016-11-12',3 union all 
select 2,'2016-11-14',7 union all 
select 2,'2016-11-15',6 union all 
select 2,'2016-11-16',3 union all 
select 2,'2016-11-17',5 union all 
select 2,'2016-11-18',3; 

-- Set the number of days that sales can accrue towards a Bonus. 
declare @DaysInBonusPeriod int = 6; 

with rn -- Derived table to get incremental ordering for recursice cte. This is useful as Sundays are ignored. 
as 
(
    select t.AgentID 
      ,t.DateValue 
      ,t.Sales 
      ,row_number() over (order by t.AgentID, t.DateValue) as rn 
    from @t t 
) 
,prev -- Using the row numbering above, find the number of sales in the day before the bonus accrual period. We have to use the row numbers as Sundays are ignored. 
as 
(
     select t.AgentID 
       ,t.DateValue 
       ,t.Sales 
       ,t.rn 
       ,isnull(tp.Sales,0) as SalesOnDayBeforeCurrentPeriod 
     from rn t 
      left join rn tp 
       on(t.AgentID = tp.AgentID 
        and tp.rn = t.rn - @DaysInBonusPeriod  -- Get number of sales on the day before the max Bonus period. 
        ) 
) 
,cte -- Use a recursive cte to calculate running totals based on sales, whether the bonus was achieved the previous day and if the previous bonus was more than 5 days ago. 
as 
(
    select rn 
      ,AgentID 
      ,DateValue 
      ,Sales 
      ,SalesOnDayBeforeCurrentPeriod 
      ,Sales as TotalSales 
      ,case when Sales >= 15 then 1 else 0 end as Bonus 
      ,1 as DaysSinceLastBonus 

    from prev 
    where rn = 1 -- Select just the first row in the dataset. 

    union all 

    select t.rn 
      ,t.AgentID 
      ,t.DateValue 
      ,t.Sales 
      ,t.SalesOnDayBeforeCurrentPeriod 

      -- If the previous row was for the same agent and not a bonus, add the day's sales to the total, subtracting the sales from the day before the 6 day bonus period if it has been more than 6 days since the last bonus. 
      ,case when t.AgentID = c.AgentID 
       then case when c.Bonus = 0 
         then t.Sales + c.TotalSales - case when c.DaysSinceLastBonus >= @DaysInBonusPeriod then t.SalesOnDayBeforeCurrentPeriod else 0 end 
         else t.Sales 
         end 
       else t.Sales 
       end as TotalSales 

      -- If the value in the TotalSales field above is 15 or more, flag a bonus. 
      ,case when 
        case when t.AgentID = c.AgentID                            --\ 
        then case when c.Bonus = 0                             -- \ 
          then t.Sales + c.TotalSales - case when c.DaysSinceLastBonus >= @DaysInBonusPeriod then t.SalesOnDayBeforeCurrentPeriod else 0 end -- \ Same statement 
          else t.Sales                              --/as TotalSales 
          end                                 --/
        else t.Sales                                --/ 
        end >= 15 
       then 1 
       else 0 
       end as Bonus 

      -- If there is no flag in Bonus field above, increment the number of days since the last bonus. 
      ,case when 
       case when                                   --\ 
         case when t.AgentID = c.AgentID                            -- \ 
         then case when c.Bonus = 0                             -- | 
           then t.Sales + c.TotalSales - case when c.DaysSinceLastBonus >= @DaysInBonusPeriod then t.SalesOnDayBeforeCurrentPeriod else 0 end -- | 
           else t.Sales                              -- \ Same statement 
           end                                 --/as Bonus 
         else t.Sales                                -- | 
         end >= 15                                 -- | 
        then 1                                   --/
        else 0                                   --/ 
        end = 0 
       then c.DaysSinceLastBonus + 1 
       else 0 
       end as DaysSinceLastBonus 

    from prev t 
     inner join cte c 
      on(t.rn = c.rn+1) 
) 
select AgentID 
     ,DateValue 
     ,Sales 
     ,TotalSales 
     ,Bonus 
from cte 
order by rn 
option (maxrecursion 0); 
+0

伟大的工作iamdave。 我认为这很复杂,看代码,甚至比我意识到更多 我得到了CTE的递归限制,但我相信我可以解决这个问题,无论是通过增加限制还是将其分解为更小卡盘。 我会对我的实时数据做更多的测试,但它看起来会起作用。 – Bob

+1

@Bob不用担心:)只需在最后添加'maxrecursion'选项,根据我的更新脚本。 “0”让它运行直到数据结束,并且任何非零值让它运行,直到递归的数量,如果它被命中则抛出一个错误。 – iamdave