从 SQL 中的日期列获取日期范围答案

【问题标题】：Get day ranges from date column in SQL从 SQL 中的日期列获取日期范围
【发布时间】：2022-01-15 04:41:38
【问题描述】：

我有一个包含日期列 (date_for) 的表，我想获取此列的连续日期范围。

select
  date_for
from (select to_date('20210910','yyyymmdd') date_for from dual union all
      select to_date('20210911','yyyymmdd') date_for from dual union all
      select to_date('20210912','yyyymmdd') date_for from dual union all
      select to_date('20210913','yyyymmdd') date_for from dual union all
      select to_date('20210915','yyyymmdd') date_for from dual union all
      select to_date('20210916','yyyymmdd') date_for from dual union all
      select to_date('20210917','yyyymmdd') date_for from dual union all
      select to_date('20211011','yyyymmdd') date_for from dual union all
      select to_date('20211012','yyyymmdd') date_for from dual union all
      select to_date('20211108','yyyymmdd') date_for from dual
      )

所需的输出类似于：

2021.09.10-2021.09.13
2021.09.15-2021.09.17
2021.10.11-2021.10.12
2021.11.08-2021.11.08

如何查询？

【问题讨论】：

这能回答你的问题吗？ Gaps and Islands Months only
顺便说一下，to_date('20210910','yyyymmdd') 可以更简单地写成date '2021-09-10'。
@William：是的，但我更喜欢使用 to_char() 函数。当您想使用参数时， to_char 很方便。它适用于世界各地，无论 NLS 设置如何。 :)
@lmocsi 不过，这些东西都不适用于这里。您不需要参数，并且 ANSI 日期文字独立于 NLS 领域。

标签： sql oracle range

【解决方案1】：

更简单的解决方案：

select
   min(date_for) st,
   max(date_for) en
from
(select
   date_for,
   date_for-rownum x
 from (select to_date('20210910','yyyymmdd') date_for from dual union all
       select to_date('20210911','yyyymmdd') date_for from dual union all
       select to_date('20210912','yyyymmdd') date_for from dual union all
       select to_date('20210913','yyyymmdd') date_for from dual union all
       select to_date('20210915','yyyymmdd') date_for from dual union all
       select to_date('20210916','yyyymmdd') date_for from dual union all
       select to_date('20210917','yyyymmdd') date_for from dual union all
       select to_date('20211011','yyyymmdd') date_for from dual union all
       select to_date('20211012','yyyymmdd') date_for from dual union all
       select to_date('20211108','yyyymmdd') date_for from dual
       )
 )
group by x
order by 1

【讨论】：

【解决方案2】：

您必须隔离岛屿。例如：

SQL> WITH
  2     test
  3     AS
  4        (SELECT TO_DATE ('20210910', 'yyyymmdd') date_for FROM DUAL
  5         UNION ALL
  6         SELECT TO_DATE ('20210911', 'yyyymmdd') date_for FROM DUAL
  7         UNION ALL
  8         SELECT TO_DATE ('20210912', 'yyyymmdd') date_for FROM DUAL
  9         UNION ALL
 10         SELECT TO_DATE ('20210913', 'yyyymmdd') date_for FROM DUAL
 11         UNION ALL
 12         SELECT TO_DATE ('20210915', 'yyyymmdd') date_for FROM DUAL
 13         UNION ALL
 14         SELECT TO_DATE ('20210916', 'yyyymmdd') date_for FROM DUAL
 15         UNION ALL
 16         SELECT TO_DATE ('20210917', 'yyyymmdd') date_for FROM DUAL
 17         UNION ALL
 18         SELECT TO_DATE ('20211011', 'yyyymmdd') date_for FROM DUAL
 19         UNION ALL
 20         SELECT TO_DATE ('20211012', 'yyyymmdd') date_for FROM DUAL
 21         UNION ALL
 22         SELECT TO_DATE ('20211108', 'yyyymmdd') date_for FROM DUAL)
 23    SELECT MIN (date_for) fie, MAX (date_for) lie
 24      FROM (SELECT date_for,
 25                     TO_NUMBER (TO_CHAR (date_for, 'yyyymmdd'))
 26                   - ROW_NUMBER () OVER (ORDER BY date_for) diff
 27              FROM test)
 28  GROUP BY diff
 29  ORDER BY 1;

FIE        LIE
---------- ----------
2021.09.10 2021.09.13
2021.09.15 2021.09.17
2021.10.11 2021.10.12
2021.11.08 2021.11.08

SQL>

在 OraFAQ 上的 Finding islands 上查看更多选项。

【讨论】：