Question

使用以下示例数据，我试图以相同的速率对记录进行分组。

   id   start_date              end_date                rate
  -----------------------------------------------------------------

    1   01/01/2017 12:00:00 am  01/01/2017 12:00:00 am  300
    1   02/01/2017 12:00:00 am  02/01/2017 12:00:00 am  300
    1   03/01/2017 12:00:00 am  03/01/2017 12:00:00 am  300
    1   04/01/2017 12:00:00 am  04/01/2017 12:00:00 am  1000
    1   05/01/2017 12:00:00 am  05/01/2017 12:00:00 am  500
    1   06/01/2017 12:00:00 am  06/01/2017 12:00:00 am  500
    1   07/01/2017 12:00:00 am  07/01/2017 12:00:00 am  1000
    1   08/01/2017 12:00:00 am  08/01/2017 12:00:00 am  1000
    1   09/01/2017 12:00:00 am  09/01/2017 12:00:00 am  300

我尝试了什么：

select distinct id, mn_date, mx_date,rate
from (
    select id, min(start_date) over (partition by grp order by start_date) mn_date,
    max(end_date) over(partition by grp order by start_date desc) mx_date, rate
    from (
    select t.*, row_number() over(partition by id order by start_date) -row_number() over(partition by rate order by start_date)grp
    from t
        )
    )
order by mn_date;

输出：

id  mn_date                 mx_date                 rate
--------------------------------------------------------
1   01/01/2017 12:00:00 am  03/01/2017 12:00:00 am  300
1   04/01/2017 12:00:00 am  04/01/2017 12:00:00 am  1000
1   05/01/2017 12:00:00 am  06/01/2017 12:00:00 am  500
1   07/01/2017 12:00:00 am  09/01/2017 12:00:00 am  300
1   07/01/2017 12:00:00 am  09/01/2017 12:00:00 am  1000

期望的输出：

id  mn_date                 mx_date                 rate
--------------------------------------------------------
1   01/01/2017 12:00:00 am  03/01/2017 12:00:00 am 300
1   04/01/2017 12:00:00 am  04/01/2017 12:00:00 am 1000
1   05/01/2017 12:00:00 am  06/01/2017 12:00:00 am 500
1   07/01/2017 12:00:00 am  08/01/2017 12:00:00 am 1000
1   09/01/2017 12:00:00 am  09/01/2017 12:00:00 am 300

按连续日期分组的最终结果:(感谢Gordon）

select id, min(start_date), max(end_date), rate
from (
select id, start_date, end_date, rate, seqnum_i-seqnum_ir grp, sum(x) over(partition by id order by start_date) grp1
from (
select t.*,
             row_number() over (partition by id order by start_date) as seqnum_i,
             row_number() over (partition by id, rate order by start_date) as seqnum_ir,
             case when LEAD(start_date) over (partition by id order by start_date)= end_date + 1 
             then 0 
             else 1
             end x
from t
)
)
group by id, grp+grp1, rate
order by min(start_date);

Answer 1

假设我们可以使用start_date来识别相邻记录（即没有间隙），那么您可以使用行号方法的差异：

select id, min(start_date) as mn_date, max(end_date) as mx_date, rate
from (select t.*,
             row_number() over (partition by id order by start_date) as seqnum_i,
             row_number() over (partition by id, rate order by start_date) as seqnum_ir
      from t
     ) t
group by id (seqnum_i - seqnum_ir), rate;

要了解其工作原理，请查看子查询的结果。你应该能够＆＃34;看到＆＃34;两个行号的差异如何定义具有相同速率的相邻记录组。

Answer 2

我发现最后一个值未正确分组，因为X的计算未处理NULL返回，因此我将其更改为：

          ,CASE
               WHEN LEAD (start_date)
                       OVER (PARTITION BY id ORDER BY start_date)
                       IS NULL
               THEN
                  0
               WHEN LEAD (start_date)
                       OVER (PARTITION BY id ORDER BY start_date) =
                       end_date + 1
               THEN
                  0
               ELSE
                  1
            END
               x

Oracle sql查询按日期对连续记录进行分组

2 个答案: