如何在 Oracle SQL 上查询以获取时间间隔，按特定字段分组

Question

我喜欢好的挑战，但这个让我头疼的时间太长了。 :)

我正在尝试构建查询以获取日期间隔，并按一个字段对信息进行分组。

让我试着用简单的方式解释一下。我们有这个 table:

我需要得到一个士兵在每个排名上花费的时间间隔，所以我需要得到的最终结果应该是这样的：

正如你所见，士兵可以 promoted/demoted 一直。

关于如何构建查询来执行此操作的任何建议？

谢谢！

Answer 1

这是一种间隙和孤岛问题。您想要找到相同的行组，您可以使用 lag() 来比较 ranking，然后使用累计和来跟踪更改：

select soldier_id, soldier_name, ranking,
       min(start_date), max(end_date)
from (select t.*,
             sum(case when prev_end_date = start_date - interval '1' day then 0 else 1 end)
                 (partition by soldier_id order by start_date) as island
      from (select t.*,
                   lag(end_date) over (partition by soldier_id, ranking order by start_date) as prev_end_date
            from t
           ) t
      ) t
group by soldier_id, soldier_name, ranking, island;

注意：这假设 soldier_name 不会随时间改变给定士兵。如果这是您需要处理的事情，请提出一个 new 问题，并提供适当的示例数据和所需的结果。

Answer 2

从 Oracle 12 开始，您可以使用 MATCH_RECOGNIZE:

SELECT *
FROM   table_name
MATCH_RECOGNIZE (
  PARTITION BY id
  ORDER     BY start_date, end_date
  MEASURES
    FIRST( name )       AS name,
    FIRST( ranking )    AS ranking,
    FIRST( start_date ) AS start_date,
    LAST( end_Date )    AS end_Date
  PATTERN ( same_rank+ )
  DEFINE same_rank AS FIRST( ranking ) = ranking
)

其中，对于示例数据：

CREATE TABLE table_name ( id, name, ranking, start_date, end_date ) AS
SELECT 1001, 'Jones', 'Lieutenant', DATE '2000-03-20', DATE '2002-08-15' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Lieutenant', DATE '2002-08-16', DATE '2003-03-18' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Lieutenant', DATE '2003-03-19', DATE '2004-06-01' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Lieutenant', DATE '2004-06-02', DATE '2004-10-01' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Captain',    DATE '2004-10-02', DATE '2005-04-20' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Captain',    DATE '2005-04-21', DATE '2007-02-20' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Major',      DATE '2007-02-21', DATE '2008-10-22' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Major',      DATE '2008-10-23', DATE '2010-01-26' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Captain',    DATE '2010-01-27', DATE '2013-11-25' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Captain',    DATE '2013-11-26', DATE '2014-05-11' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'Major',      DATE '2014-05-12', DATE '2016-04-22' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'General',    DATE '2016-04-23', DATE '2020-10-10' FROM DUAL UNION ALL
SELECT 1001, 'Jones', 'General',    DATE '2020-10-11', DATE '2020-11-30' FROM DUAL;

输出：

  ID | NAME  | RANKING    | START_DATE          | END_DATE           
---: | :---- | :--------- | :------------------ | :------------------
1001 | Jones | Lieutenant | 2000-03-20 00:00:00 | 2004-10-01 00:00:00
1001 | Jones | Captain    | 2004-10-02 00:00:00 | 2007-02-20 00:00:00
1001 | Jones | Major      | 2007-02-21 00:00:00 | 2010-01-26 00:00:00
1001 | Jones | Captain    | 2010-01-27 00:00:00 | 2014-05-11 00:00:00
1001 | Jones | Major      | 2014-05-12 00:00:00 | 2016-04-22 00:00:00
1001 | Jones | General    | 2016-04-23 00:00:00 | 2020-11-30 00:00:00

db<>fiddle here

如何在 Oracle SQL 上查询以获取时间间隔，按特定字段分组

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