Question

我试图获得每个组的最小和最大行数，结束日期与下一行开始日期匹配

输入日期：

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

期望的结果：

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

我试过了：

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

您可以通过检查上一个结束日期来确定组的开始位置。然后，可以通过累积和来为相邻日期组分配组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;

按之前的日期值分组

1 个答案: