SQL查詢顯示連續3行或更多行且人數超過100人的記錄

Question

我有一個具有以下結構的表：

+----+------------+--------+
| id | visit_date | people |
+----+------------+--------+
|  1 | 2017-01-01 |     10 |
|  2 | 2017-01-02 |    109 |
|  3 | 2017-01-03 |    150 |
|  4 | 2017-01-04 |     99 |
|  5 | 2017-01-05 |    145 |
|  6 | 2017-01-06 |   1455 |
|  7 | 2017-01-07 |    199 |
|  8 | 2017-01-08 |    188 |
+----+------------+--------+

我需要一個查詢來顯示連續 3 行或更多行且人數超過 100 的記錄。所以輸出將是：

+----+------------+--------+
| id | visit_date | people |
+----+------------+--------+
|  5 | 2017-01-05 |    145 |
|  6 | 2017-01-06 |   1455 |
|  7 | 2017-01-07 |    199 |
|  8 | 2017-01-08 |     88 |
+----+------------+--------+

為方便起見，創建表語句：

CREATE TABLE stadium(id INT, visit_date DATE, people INT);
INSERT INTO stadium VALUES(1, '01/JAN/17', 10);
INSERT INTO stadium VALUES(1, '02/JAN/17', 109);
INSERT INTO stadium VALUES(1, '03/JAN/17', 150);
INSERT INTO stadium VALUES(1, '04/JAN/17', 99);
INSERT INTO stadium VALUES(1, '05/JAN/17', 145);
INSERT INTO stadium VALUES(1, '06/JAN/17', 1455);
INSERT INTO stadium VALUES(1, '07/JAN/17', 199);
INSERT INTO stadium VALUES(1, '08/JAN/17', 188);

這是在 Oracle SQL 中

如果有人可以解釋邏輯和解決方案，我將非常感激。

Answer 1

這是一種間隙和島嶼問題。 您可以使用行號的差異來識別組。 然后使用count(*)來計算行數和過濾器：

select s.*
from (select s.*,
             count(*) over (partition by id, seqnum - seqnum_2) as num_in_row
      from (select s.*,
                   row_number() over (partition by id order by visit_date) as seqnum,
                   row_number() over (partition by id, (case when people > 100 then 1 else 0 end) order by visit_date) as seqnum_2
            from stadium s
           ) s
      where people > 100
     ) s
where num_in_row >= 3;

這是 SQL 小提琴。

Answer 2

這是一個典型的模式匹配問題。 match_recognize可以快速完成此類任務。

select id, visit_date, people
from   stadium
match_recognize(
  order by visit_date
  all rows per match
  pattern ( a{3, } )
  define a as people >= 100    --  or > 100?
);