2016-11-09 71 views
3

我正在使用SQL Server 2012爲銷售代理提取滾動銷售信息。 如果這些代理商在6天或更少的滾動時間內達到15次銷售額,則會獲得獎金。如果他們成爲目標,則滾動計數會重置。 週日應該被忽略。銷售目標使用重置的運行總計SQL Server 2012

因此,考慮以下AGENTID,日期和銷售數據:

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 

,我期望的目標被擊中的日期是:

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 

我已經嘗試了幾種方法,但我找不到重置滾動總數的方法。

我認爲窗口函數是要走的路。

裁減員額像 Window Functions - Running Total with reset

我認爲這是類似於我所需要的,但不能完全得到它才能正常工作。

更新: 我嘗試的第一件事是創建滾動6天的窗口,但我沒有看到這在一個基於集合的方法工作。 我可以使用光標來遍歷這些行,但我真的不喜歡這個想法。

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 

然後我試圖用在上面的URL中提到的腳本,但我真的不知道它在做什麼。 我只是在改變一些東西,希望在解決方案中陷入困境,而這只是不起作用。

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

你有你至今嘗試過的腳本? – iamdave

+1

另外,同一天15日之後的銷售額是否會計入下一個目標?因此,如果他們在第1天售出14,然後在第2天售出3,那麼對第3天的第2個或第2個目標開始計數? – iamdave

+0

哦,還有(!)是否每個代理每天最多隻有一條記錄? – iamdave

回答

1

那就好!這是一個有趣的挑戰,我非常高興,因爲我破解了它。註釋等在代碼註釋中。如果您想更改可累積獎金的天數,請更改@DaysInBonusPeriod中的值。這也適用於多個AgentID秒和日期的任何序列,假設其中的任何缺少的日期不包括獎金計提時期 - 即:如果你忽視週日週三時段進行計數:

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

解決方案

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

偉大的工作iamdave。 我認爲這很複雜,看代碼,甚至比我意識到更多 我得到了CTE的遞歸限制,但我相信我可以解決這個問題,無論是通過增加限制還是將其分解爲更小卡盤。 我會對我的實時數據做更多的測試,但它看起來會起作用。 – Bob

+1

@Bob不用擔心:)只需在最後添加'maxrecursion'選項,根據我的更新腳本。 「0」讓它運行直到數據結束,並且任何非零值讓它運行,直到遞歸的數量,如果它被命中則拋出一個錯誤。 – iamdave