SQL查询可在特定部件修复后查找最多3笔交易

Question

我的数据与此示例非常相似，因此非常大，因此idealy需要高效的代码。我想在充电器修理后找到这些交易（最多3个）。

TRANSACTION_ID REPAIR_DATE CUSTOMER_ID COMPONENT LABOR_CODE_DESC .... 
28289   6/25/2015 AH123  LAPTOP  CHARGER REPAIR 
28235   6/29/2015 AH123  LAPTOP   CHIP REPLACE 
258978   6/27/2013  HW687  PHONE  TOUCH SCREEN 
28223   6/2/2014  AH123  LAPTOP  BATTERY REPAIR 
215678   9/7/2014  HW687  PHONE   SIM REPAIR 
527808   7/30/2016  HW687  LAPTOP   BATTERY REPAIR 
567976   7/28/2014  HW687  LAPTOP   CHARGER REPAIR 
7678698   8/68/2015  AH123  LAPTOP  BATTERY REPAIR 
9987908   5/7/2006  TU890  PHONE   SIM REPAIR 
..... 


OUTPUT 
TRANSACTION_ID REPAIR_DATE CUSTOMER_ID COMPONENT LABOR_CODE_DESC .... 
28235   6/29/2015 AH123  LAPTOP   CHIP REPLACE 
7678698   8/68/2015  AH123  LAPTOP  BATTERY REPAIR 
527808   7/30/2016  HW687  LAPTOP   BATTERY REPAIR 
215678   9/7/2014  HW687  PHONE  SIM REPAIR 

..... 

因此，所有我想要的是那些客户ID的交易已经过充电修复和充电器维修日期

SELECT TRANSACTION_ID, REPAIR_DATE,CUSTOMER_ID,LABOR_CODE_DESC from table 
where customer_id IN (SELECT CUSTOMER_ID from table where LABOR_CODE_DESC 
like '%CHARGER REPAIR%') 

不知道我该怎么扩大，以获得最大的3后最多3个交易充电修复后交易

Answer 1

这是一个充电修复日期后返回三行：

SELECT * 
FROM tab 
QUALIFY 
    -- check if any of the three previous rows contains 'CHARGER REPAIR' 
    Max(CASE WHEN LABOR_CODE_DESC = 'CHARGER REPAIR' THEN 1 ELSE 0 END) 
    Over (PARTITION BY CUSTOMER_ID 
     ORDER BY REPAIR_DATE 
     ROWS BETWEEN 3 Preceding AND 1 Preceding) = 1 

编辑：

要在未来10天内仅获得具有修理日期的行：

-- check if any of the previous rows contains 'CHARGER REPAIR' 
    -- and the current REPAIR_DATE is within 10 days after that repair 
    Max(CASE WHEN LABOR_CODE_DESC = 'CHARGER REPAIR' THEN REPAIR_DATE END) 
    Over (PARTITION BY CUSTOMER_ID 
     ORDER BY REPAIR_DATE 
     ROWS BETWEEN Unbounded Preceding AND 1 Preceding) >= REPAIR_DATE - 10 

Answer 2

您可以通过执行获得所有此类交易：

select t.* 
from table t 
where t.repair_date > (select t2.repair_date 
         from table t2 
         where t2.customer_id = t.customer_id and 
          labor_code_desc like '%CHARGER REPAIR%' 
        ); 

限制这3％的客户是小tricker，但您可以使用窗口函数：

select t.* 
from (select t.*, 
      row_number() over (partition by customer_id order by repair_date) as seqnum 
     from (select t.*, 
        max(case when labor_code_desc like '%CHARGER REPAIR%' then repair_date end) over (partition by customer_id) as max_crd 
      from table t 
      ) t 
     where t.repair_date > max_crd 
    ) t 
where seqnum <= 3;