数一个岛及其起点和终点

Question

我有一个 table 看起来像这样：

我想要的是根据 Day_Ranking 列计算数据孤岛，因此输出看起来类似于：

Person           Start Date           End Date           Count
 45286            2021-08-26           2021-09-07         8
 13914            2021-09-22           2021-10-01         7
 32914            2021-01-10           2021-01-10         1
 32914            2021-01-12           2021-01-13         2

有什么关于如何做到这一点的建议吗？ 我正在使用 Oracle SQL

Answer 1

从 Oracle 12 开始，您可以使用 MATCH_RECOGNIZE 执行 row-by-row 比较：

SELECT *
FROM   table_name
MATCH_RECOGNIZE(
  PARTITION BY person
  ORDER BY day_ranking
  MEASURES
    FIRST(date_value) AS start_date,
    LAST(date_value) AS end_date,
    COUNT(*) AS count
  ONE ROW PER MATCH
  PATTERN (successive_ranks* last_rank)
  DEFINE successive_ranks AS day_ranking + 1 = NEXT(day_ranking)
)

其中，对于示例数据：

CREATE TABLE table_name (person, date_value, day_ranking) AS
SELECT 1, DATE '2022-01-01',  6 FROM DUAL UNION ALL
SELECT 1, DATE '2022-01-02',  7 FROM DUAL UNION ALL
SELECT 1, DATE '2022-01-04',  8 FROM DUAL UNION ALL
SELECT 1, DATE '2022-01-05',  9 FROM DUAL UNION ALL
SELECT 1, DATE '2022-01-07', 10 FROM DUAL UNION ALL
SELECT 1, DATE '2022-01-09', 11 FROM DUAL UNION ALL
SELECT 2, DATE '2022-01-01', 15 FROM DUAL UNION ALL
SELECT 2, DATE '2022-01-02', 16 FROM DUAL UNION ALL
SELECT 2, DATE '2022-01-03', 17 FROM DUAL UNION ALL
SELECT 2, DATE '2022-01-04', 18 FROM DUAL UNION ALL
SELECT 2, DATE '2022-01-05', 19 FROM DUAL UNION ALL
SELECT 3, DATE '2022-01-01', 25 FROM DUAL UNION ALL
SELECT 3, DATE '2022-01-03', 27 FROM DUAL;

输出：

PERSON	START_DATE	END_DATE	COUNT
1	01-JAN-22	09-JAN-22	6
2	01-JAN-22	05-JAN-22	5
3	01-JAN-22	01-JAN-22	1
3	03-JAN-22	03-JAN-22	1

db<>fiddle here