5

在工作中,我的一项任务是计算销售人员的佣金。其中一条规则比其他规则更具挑战性。使用递归CTE处理日期逻辑

两个销售团队A和B一起工作,每个销售不同的产品。 A队可以将线索发送给B队。同一个客户可能会多次发送。第一次将客户(如主管1)*发送佣金时支付给创建主管的团队A中的销售人员。现在,客户在未来365天内(从创建导致1的日期算起)“锁定”。这意味着在那段时间内没有人可以通过发送额外的线索为该客户获得额外的佣金(例如,Lead 2和3不获得佣金)。 365天过期后。新的潜在客户可以创建并获得佣金(例如,潜在客户4)。然后,客户再次被锁定,从铅4创建日起计算365天。因此,领导5不获得任何佣金。棘手的部分是重置365天计算的日期。

'*引用表#LEADS和#DISERED结果。

我已经解决了使用游标在tSQL中的问题,但我想知道是否有可能使用递归CTE。我已经做了几次尝试,最好的贴在下面。我的解决方案的问题是,我不止一次地引用递归表。我试图通过在CTE中嵌套CTE来解决这个问题。这是不允许的。我曾尝试在CTE内部使用临时表,这是不允许的。我尝试了多次重新编码CTE的递归部分,以便递归表只被引用一次,但后来我无法使逻辑工作。

我使用SQL 2008

IF OBJECT_ID('tempdb.dbo.#LEADS', 'U') IS NOT NULL 
DROP TABLE #LEADS; 

CREATE TABLE #LEADS (LEAD_ID INT, CUSTOMER_ID INT, LEAD_CREATED_DATE DATETIME, SALESPERSON_NAME varchar(20)) 
INSERT INTO #LEADS 
VALUES (1, 1, '2013-09-01', 'Rasmus') 
     ,(2, 1, '2013-11-01', 'Christian') 
     ,(3, 1, '2014-01-01', 'Nadja') 
     ,(4, 1, '2014-12-24', 'Roar') 
     ,(5, 1, '2015-12-01', 'Kristian') 
     ,(6, 2, '2014-01-05', 'Knud') 
     ,(7, 2, '2015-01-02', 'Rasmus') 
     ,(8, 2, '2015-01-08', 'Roar') 
     ,(9, 2, '2016-02-05', 'Kristian') 
     ,(10, 2, '2016-03-05', 'Casper') 

SELECT * 
FROM #LEADS; 

IF OBJECT_ID('tempdb.dbo.#DISERED_RESULT', 'U') IS NOT NULL 
DROP TABLE #DISERED_RESULT; 

CREATE TABLE #DISERED_RESULT (LEAD_ID INT, DESIRED_COMMISION_RESULT CHAR(3)) 
INSERT INTO #DISERED_RESULT 
VALUES (1, 'YES') 
     ,(2, 'NO') 
     ,(3, 'NO') 
     ,(4, 'YES') 
     ,(5, 'NO') 
     ,(6, 'YES') 
     ,(7, 'NO') 
     ,(8, 'YES') 
     ,(9, 'YES') 
     ,(10, 'NO') 

    SELECT * 
    FROM #DISERED_RESULT; 

WITH COMMISSION_CALCULATION AS 
(
    SELECT T1.* 
      ,COMMISSION = 'YES' 
      ,MIN_LEAD_CREATED_DATE AS COMMISSION_DATE 
    FROM  #LEADS AS T1 
    INNER JOIN (
       SELECT A.CUSTOMER_ID 
         ,MIN(A.LEAD_CREATED_DATE) AS MIN_LEAD_CREATED_DATE 
       FROM #LEADS AS A 
       GROUP BY A.CUSTOMER_ID 
       ) AS T2 ON T1.CUSTOMER_ID = T2.CUSTOMER_ID AND T1.LEAD_CREATED_DATE = T2.MIN_LEAD_CREATED_DATE 

UNION ALL 

SELECT T10.LEAD_ID 
     ,T10.CUSTOMER_ID 
     ,T10.LEAD_CREATED_DATE 
     ,T10.SALESPERSON_NAME 
     ,T10.COMMISSION 
     ,T10.COMMISSION_DATE 
FROM (SELECT ROW_NUMBER() OVER(PARTITION BY T5.CUSTOMER_ID ORDER BY T5.LEAD_CREATED_DATE ASC) AS RN 
       ,T5.* 
       ,T6.MAX_COMMISSION_DATE 
       ,DATEDIFF(DAY, T6.MAX_COMMISSION_DATE, T5.LEAD_CREATED_DATE) AS ANTAL_DAGE_SIDEN_SIDSTE_COMMISSION 
       ,CASE 
        WHEN DATEDIFF(DAY, T6.MAX_COMMISSION_DATE, T5.LEAD_CREATED_DATE) > 365  THEN 'YES' 
        ELSE 'NO' 
       END AS COMMISSION 
       ,CASE 
        WHEN DATEDIFF(DAY, T6.MAX_COMMISSION_DATE, T5.LEAD_CREATED_DATE) > 365  THEN T5.LEAD_CREATED_DATE 
        ELSE NULL 
       END AS COMMISSION_DATE 
     FROM  #LEADS AS T5 
      INNER JOIN (SELECT  T4.CUSTOMER_ID 
            ,MAX(T4.COMMISSION_DATE) AS MAX_COMMISSION_DATE 
         FROM  COMMISSION_CALCULATION AS T4 
         GROUP BY T4.CUSTOMER_ID) AS T6 ON T5.CUSTOMER_ID = T6.CUSTOMER_ID 
     WHERE T5.LEAD_ID NOT IN (SELECT LEAD_ID FROM COMMISSION_CALCULATION) 
     ) AS T10 
WHERE RN = 1 


) 
    SELECT * 
    FROM COMMISSION_CALCULATION; 
+1

加1的样本数据,请张贴您的预期结果 – TheGameiswar

+0

样本数据的最后一列是期望的结果。感谢您的时间。 –

+0

您的查询有一些问题,你可以在问题中粘贴预期结果 – TheGameiswar

回答

0

我已经取得了一些假设您的描述不完全意义的写,但低于达到您想要的结果:

if object_id('tempdb.dbo.#leads', 'u') is not null 
drop table #leads; 

create table #leads (lead_id int, customer_id int, lead_created_date datetime, salesperson_name varchar(20)) 
insert into #leads 
values (1, 1, '2013-09-01', 'rasmus') 
     ,(2, 1, '2013-11-01', 'christian') 
     ,(3, 1, '2014-01-01', 'nadja') 
     ,(4, 1, '2014-12-24', 'roar') 
     ,(5, 1, '2015-12-01', 'kristian') 
     ,(6, 2, '2014-01-05', 'knud') 
     ,(7, 2, '2015-01-02', 'rasmus') 
     ,(8, 2, '2015-01-08', 'roar') 
     ,(9, 2, '2016-02-05', 'kristian') 
     ,(10, 2, '2016-03-05', 'casper') 

if object_id('tempdb.dbo.#disered_result', 'u') is not null 
drop table #disered_result; 

create table #disered_result (lead_id int, desired_commision_result char(3)) 
insert into #disered_result 
values (1, 'yes'),(2, 'no'),(3, 'no'),(4, 'yes'),(5, 'no'),(6, 'yes'),(7, 'no'),(8, 'yes'),(9, 'yes'),(10, 'no') 

with rownum 
as 
(
    select row_number() over (order by customer_id, lead_created_date) as rn        -- This is to ensure an incremantal ordering id 
      ,lead_id 
      ,customer_id 
      ,lead_created_date 
      ,salesperson_name 
    from #leads 
) 
,cte 
as 
(
    select rn 
      ,lead_id 
      ,customer_id 
      ,lead_created_date 
      ,salesperson_name 
      ,'yes' as commission_result 
      ,lead_created_date as commission_window_start 
    from rownum 
    where rn = 1 

    union all 

    select r.rn 
      ,r.lead_id 
      ,r.customer_id 
      ,r.lead_created_date 
      ,r.salesperson_name 

      ,case when r.customer_id <> c.customer_id  -- If the customer ID has changed, we are at a new commission window. 
       then 'yes' 
       else case when r.lead_created_date > dateadd(d,365,c.commission_window_start) -- This assumes the window is 365 days and not one year (ie. Leap years don't add a day) 
         then 'yes' 
         else 'no' 
         end 
       end as commission_result 

      ,case when r.customer_id <> c.customer_id 
       then r.lead_created_date 
       else case when r.lead_created_date > dateadd(d,365,c.commission_window_start) -- This assumes the window is 365 days and not one year (ie. Leap years don't add a day) 
         then r.lead_created_date 
         else c.commission_window_start 
         end 
       end as commission_window_start 

    from rownum r 
     inner join cte c 
      on(r.rn = c.rn+1) 
) 
select lead_id 
     ,commission_result 
from cte 
order by customer_id 
     ,lead_created_date; 
+0

感谢您花时间制作解决方案。该解决方案提供了期望的结果。我期待在我的佣金计算项目中实现它,以便了解性能如何与基于光标的解决方案进行比较。 –