5

在工作中,我的一項任務是計算銷售人員的佣金。其中一條規則比其他規則更具挑戰性。使用遞歸CTE處理日期邏輯

兩個銷售團隊A和B一起工作,每個銷售不同的產品。 A隊可以將線索發送給B隊。同一個客戶可能會多次發送。第一次將客戶(如主管1)*發送佣金時支付給創建主管的團隊A中的銷售人員。現在,客戶在未來365天內(從創建導致1的日期算起)「鎖定」。這意味着在那段時間內沒有人可以通過發送額外的線索爲該客戶獲得額外的佣金(例如,Lead 2和3不獲得佣金)。 365天過期後。新的潛在客戶可以創建並獲得佣金(例如,潛在客戶4)。然後,客戶再次被鎖定,從鉛4創建日起計算365天。因此,領導5不獲得任何佣金。棘手的部分是重置365天計算的日期。

'*引用表#LEADS和#DISERED結果。

我已經解決了使用遊標在tSQL中的問題,但我想知道是否有可能使用遞歸CTE。我已經做了幾次嘗試,最好的貼在下面。我的解決方案的問題是,我不止一次地引用遞歸表。我試圖通過在CTE中嵌套CTE來解決這個問題。這是不允許的。我曾嘗試在CTE內部使用臨時表,這是不允許的。我嘗試了多次重新編碼CTE的遞歸部分,以便遞歸表只被引用一次,但後來我無法使邏輯工作。

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的樣本數據,請張貼您的預期結果 – TheGameiswar

+0

樣本數據的最後一列是期望的結果。感謝您的時間。 –

+0

您的查詢有一些問題,你可以在問題中粘貼預期結果 – TheGameiswar

回答

0

我已經取得了一些假設您的描述不完全意義的寫,但低於達到您想要的結果:

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

感謝您花時間製作解決方案。該解決方案提供了期望的結果。我期待在我的佣金計算項目中實現它,以便了解性能如何與基於光標的解決方案進行比較。 –