集團由以前的日期值

Question

我試圖讓每組的最小值和最大值的行相匹配的下一行開始日期

輸入一個日期結束日期：

ResultUid BeginDate     EndDate 
1   1999-12-31 00:00:00.000 2000-01-31 00:00:00.000 
1   2000-01-31 00:00:00.000 2000-02-29 00:00:00.000 
1   2000-02-29 00:00:00.000 2000-03-31 00:00:00.000 
1   2000-03-31 00:00:00.000 2000-04-30 00:00:00.000 
1   2007-03-31 00:00:00.000 2007-04-30 00:00:00.000 
1   2007-04-30 00:00:00.000 2007-05-31 00:00:00.000 
1   2007-05-31 00:00:00.000 2007-06-30 00:00:00.000 

所需的結果：

ResultUid BeginDate     EndDate 
1   1999-12-31 00:00:00.000 2000-04-30 00:00:00.000 
1   2007-03-31 00:00:00.000 2007-06-30 00:00:00.000 

我已經試過：

SELECT 
    ResultUid, 
    MIN(BeginDate) AS "min", 
    MAX(EndDate) AS "max", 
    lag 
FROM (
    SELECT 
     ResultUid, 
     BeginDate, 
     EndDate, 
     DATEDIFF(MONTH,lag(BeginDate) OVER (order by EndDate), EndDate) AS "lag" 
    FROM Results 
    GROUP BY 
     ResultUid, 
     BeginDate, 
     EndDate 
) sub 
GROUP BY 
    ResultUid, 
    lag 

Answer 1

您可以通過檢查上一個結束日期來確定組的起始位置。然後，相鄰日期的組可以通過累積總和來分配組ID：

select resultuid, min(begindate) as begindate, max(enddate) as enddate 
from (select r.*, 
      sum(case when prev_enddate = begindate then 0 else 1 end) over 
       (partition by resultuid order by begindate) as grp 
     from (select r.*, 
        lag(enddate) over (partition by resultuid order by begindate) as prev_enddate 
      from results r 
      ) r 
    ) r 
group by resultuid, grp;