如何在查找日期之间的重叠时查找哪些行正在重叠

Question

我们如何找到日期之间的重叠圈。重叠意味着当开始日期和结束日期在以下示例的相同范围内时，第 1 行没有重叠。第 2 到 5 行可以被视为一组重叠，因为开始日期和结束日期与它们自己重叠第 6 行和第 7 行可以被视为一组重叠，例如。第 6 行和第 7 行 --> 第 7 行的开始日期与第 6 行的结束日期在同一范围内，因此它成为重叠

一旦发现重叠，则需要找出 min(start date) 和 max(end date) 和 想要为每个重叠分配一个唯一的 id，并且在 S.NO 列中应该显示哪些行重叠。

下面是 I/p 和 O/p

Answer 1

基于this answer（其中包含查询的详细说明），您可以更改最后部分以生成重叠行组，然后使用LISTAGG聚合以获得分隔的sno值：

SELECT id,
       MIN( dt ) AS start_date,
       MAX( dt ) AS end_date,
       LISTAGG( CASE value WHEN 1 THEN sno END, ',' )
         WITHIN GROUP ( ORDER BY dt ) AS snos
FROM   (
  SELECT sno,
         id,
         dt,
         value,
         SUM( start_end ) OVER ( ORDER BY dt ASC, value DESC )
           AS grp
  FROM   (
    SELECT sno,
           id,
           dt,
           value,
           CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
           WHEN 1 THEN 1
           WHEN 0 THEN 0
           ELSE NULL
           END AS start_end
    FROM   table_name
    UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
  )
)
GROUP BY id, grp;

其中，对于样本数据：

CREATE TABLE table_name ( sno, id, start_date, end_date ) AS
SELECT 1, 1, DATE '2020-10-11', DATE '2020-10-11' FROM DUAL UNION ALL
SELECT 2, 1, DATE '2020-11-04', DATE '2020-12-11' FROM DUAL UNION ALL
SELECT 3, 1, DATE '2020-11-05', DATE '2020-11-10' FROM DUAL UNION ALL
SELECT 4, 1, DATE '2020-11-06', DATE '2020-11-10' FROM DUAL UNION ALL
SELECT 5, 1, DATE '2020-11-20', DATE '2020-12-20' FROM DUAL UNION ALL
SELECT 6, 1, DATE '2021-01-01', DATE '2021-01-20' FROM DUAL UNION ALL
SELECT 7, 1, DATE '2021-01-01', DATE '2021-03-25' FROM DUAL;

输出：

身份证 | START_DATE | END_DATE | SNOS -: | :----------------- | :----------------- | :------ 1 | 2020-10-11 00:00:00 | 2020-10-11 00:00:00 | 1 1 | 2020-11-04 00:00:00 | 2020-12-20 00:00:00 | 2,3,4,5 1 | 2021-01-01 00:00:00 | 2021-03-25 00:00:00 | 6,7

db小提琴here

---

从 Oracle 12c 开始，您可以使用 MATCH_RECOGNIZE 进行更简单的查询：

SELECT id,
       MIN( start_date ) AS start_date,
       MAX( end_date )   AS end_date,
       LISTAGG( sno, ',' ) WITHIN GROUP ( ORDER BY start_date ASC, END_DATE ASC )
         AS snos
FROM   table_name
MATCH_RECOGNIZE (
  PARTITION BY id
  ORDER     BY start_date ASC, end_date ASC
  MEASURES  MATCH_NUMBER() AS grp
  ALL ROWS PER MATCH
  PATTERN ( OVERLAPPING_DATES* LAST_DATE )
  DEFINE OVERLAPPING_DATES AS NEXT( start_date ) <= MAX( end_date )
)
GROUP BY id, grp

db小提琴here

Answer 2

这是一种孤岛问题。您可以使用累积最大值来确定是否存在重叠来解决此问题。然后累积总和和聚合做你想要的：

select id, min(start_date), max(end_date),
       listagg(s_no, ',') within group (order by s_no) as s_nos
from (select t.*,
             sum(case when prev_end_date >= start_date then 0 else 1 end) over (partition by id order by s_no) as grp
      from (select t.*,
                   max(end_date) over (
                       partition by id
                       order by s_no
                       rows between unbounded preceding and 1 preceding
                      ) as prev_end_date
            from t
           ) t
     ) t
group by id, grp;

Here 是一个 dbfiddle。