Oracle sql查询按日期对连续记录进行分组

使用下面的示例数据，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

我所尝试的：

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）

---

假设我们可以使用

start\u date

来识别相邻的记录（即没有间隙），那么您可以使用行号差异法：

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。您应该能够“看到”两个行号的差异如何定义具有相同速率的相邻记录组。

我发现最后一个值没有正确分组，因为X的计算没有处理空返回，所以我将其更改为：

   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

          ,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

---

您希望如何对数据进行分组？输出和期望输出之间的区别是什么？你尝试了什么？谢谢戈登：），我第一次尝试的查询也是基于你在另一个线程中的输入。谢谢你的新解决方案。如果我想将上述记录作为连续日期范围的一个组，你能建议更改吗？假设我有一个额外条目，现有列表的开始日期和结束日期分别为（2017年1月11日至2017年1月13日），费率为300。查询结果应该是新行，因为它不是连续的日期？我已尝试使用lead（），但未能获得预期的结果。还有一个条目（2017年1月15日和2017年1月15日），费率300。