Question

我有一个表，该表在一定时间后会获得陈旧（相似）值。我想找到最后一个连续值的计数。您能帮我为此编写一个Oracle SQL查询吗？我已经在下面打印了示例输入和输出。

输入：

ID   date         value
2641 02/01/2019    5.6
2641 02/04/2019    5.7
2641 02/05/2019    6.2
2641 02/06/2019    6.2
2641 02/07/2019    4.3 
2641 02/08/2019    4.3
2641 02/11/2019    4.3
2641 02/12/2019    4.3
2641 02/13/2019    4.3
2641 02/14/2019    4.3
2641 02/15/2019    4.3
2641 02/18/2019    4.3
2641 02/19/2019    4.3
2642 02/01/2019    7.6
2642 02/04/2019    4.7
2642 02/05/2019    5.2
2642 02/06/2019    3.8
2642 02/07/2019    3.8
2642 02/08/2019    3.8
2642 02/11/2019    3.8
2642 02/12/2019    3.8
2642 02/13/2019    3.8
2642 02/14/2019    3.8
2642 02/15/2019    3.8
2642 02/18/2019    3.8
2642 02/19/2019    3.8

输出

ID   stale_days
2641  9  
2642  10

关于，沙旺

Answer 1

我会那样做。

想法是让每一行都获得后面所有行的最大值和后面所有行的最小值。如果该值与最小值和最大值相同，则将它们计数。

SELECT ID, COUNT(*) FROM (
  SELECT 
    ID, 
    VALUE,
    MIN(VALUE) OVER ( PARTITION BY ID ORDER BY D ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING ) AS MIN_NEXT,
    MAX(VALUE) OVER ( PARTITION BY ID ORDER BY D ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING ) AS MAX_NEXT
  FROM T
) WHERE VALUE=MIN_NEXT AND VALUE=MAX_NEXT
GROUP BY ID;

对于测试：http://sqlfiddle.com/#!4/046b1/1

Answer 2

    with rows_with_cons_val as
    (
    select t.*, rowid, row_number() over (partition by id, value order by date_entered) 
    rid 
    from t) 
    select id, rid as stale_days
    from (
    select r.*, max(date_entered) over (partition by id) as selection
    from rows_with_cons_val r) where date_entered = selection;

查找列中连续相似值的数量-Oracle

2 个答案: