制作一条记录，将其重复记录基于两列的组合并分配 Min start_date 和 Max end_date

Question

这张图片将向您展示问题结构

这应该是所需的输出

我想填充 Min Start_Date 和 Max End_Date 来自该记录的重复记录（基于 Present 和 Absent 列的重复）和制作主记录，与 Present 和 Absent 列的其他组合相同。

我已经根据 ID 和开始日期完成排序，以了解数据的行为。这个场景我只提到了一个 ID，其他 ID 也应该给出相同类型的输出，如果我得到这个例子的有效解决方案，我可以在整个 table 上实现正确的逻辑。我试过使用 window 函数但没有找到任何解决方案。提前致谢

请注意 ID 的重复记录数是可变的。

Answer 1

这是一个 gaps-and-islands 问题。考虑使用行号之间的差异来构建“相邻”记录组，然后您可以合并：

select id, min(start_date) start_date, max(end_date end_date, present, absent
from (
    select t.*,
        row_number() over(partition by id order by start_date) rn1,
        row_number() over(partition by id, present, absent order by start_date) rn2
    from mytable t
) t
group by id, present, absent, rn1 - rn2
order by 1, 2

Answer 2

您可以使用 MATCH_RECOGNIZE:

SELECT *
FROM   table_name
MATCH_RECOGNIZE (
   PARTITION BY id
   ORDER BY     start_date
   MEASURES     FIRST( start_date ) AS start_date,
                MAX( end_date )     AS end_date,
                FIRST( present )    AS present,
                FIRST( absent )     AS absent
   ONE ROW PER MATCH
   PATTERN      (FIRST_ROW EQUAL_ROWS*)
   DEFINE       EQUAL_ROWS AS
   (
     (
       (
         EQUAL_ROWS.present = PREV(EQUAL_ROWS.present)
       ) OR (
         EQUAL_ROWS.present IS NULL AND PREV(EQUAL_ROWS.present) IS NULL
       )
     ) AND (
       (
         EQUAL_ROWS.absent  = PREV(EQUAL_ROWS.absent)
       ) OR (
         EQUAL_ROWS.absent IS NULL AND PREV(EQUAL_ROWS.absent) IS NULL
       )
     )
   )
)

因此，对于您的示例数据：

CREATE TABLE table_name ( id, start_date, end_date, present, absent ) AS
SELECT 1, DATE '2020-02-01', DATE '2020-03-01', 'Y',  'N'  FROM DUAL UNION ALL
SELECT 1, DATE '2020-03-04', DATE '2020-04-19', 'Y',  'N'  FROM DUAL UNION ALL
SELECT 1, DATE '2020-03-06', DATE '2020-03-09', 'N',  'N'  FROM DUAL UNION ALL
SELECT 1, DATE '2020-05-04', DATE '2020-09-04', 'N',  'Y'  FROM DUAL UNION ALL
SELECT 1, DATE '2020-05-06', DATE '2020-06-26', 'N',  'Y'  FROM DUAL UNION ALL
SELECT 1, DATE '2020-07-12', DATE '2020-08-12', NULL, NULL FROM DUAL UNION ALL
SELECT 1, DATE '2020-08-13', DATE '2020-08-12', NULL, NULL FROM DUAL;

这输出：

ID | START_DATE | END_DATE  | PRESENT | ABSENT
-: | :--------- | :-------- | :------ | :-----
 1 | 01-FEB-20  | 19-APR-20 | Y       | N     
 1 | 06-MAR-20  | 09-MAR-20 | N       | N     
 1 | 04-MAY-20  | 04-SEP-20 | N       | Y     
 1 | 12-JUL-20  | 12-AUG-20 | null    | null

db<>fiddle here

制作一条记录，将其重复记录基于两列的组合并分配 Min start_date 和 Max end_date

Make a record combining its duplicate records based combination of two columns and assign the Min start_date and Max end_date

sql

oracle

gaps-and-islands

db2-luw