基于列序列的间隙和孤岛查询/重置行数

Posted 2023-03-31

【中文标题】基于列序列的间隙和孤岛查询/重置行数

【英文标题】：Gaps and Islands query / reset row count based on column seqeunce

【发布时间】：2016-04-20 09:00:52

【问题描述】：

我知道有很多这样的例子，但仍然找不到确切的解决方案。

我希望根据 1 和 0 的序列重置行数。

DECLARE @TestTable TABLE (category INT, ts INT,window int)
INSERT INTO @TestTable (category,ts,window)
VALUES  (1,1,1),(1,2,1),(1,3,0),(1,4,0),(1,5,1),(1,6,1),(1,7,1),(2,1,0),(2,2,1),(2,3,1),(2,4,1),(2,5,0),(2,6,0),(2,7,1),(2,8,1),(2,9,1),(2,10,1),(2,11,1)

在上表中，我想要一个递增 1 的行号列，按类别分区，但每次窗口更改时计数都需要重置。

到目前为止，我得到的最好的结果是：

SELECT
    x.category,
    ts,
    window, 
    is_group, 
    SUM( is_group ) OVER (PARTITION BY x.category ORDER BY ts ROWS BETWEEN 1 PRECEDING AND 0 FOLLOWING ) * is_group
FROM
    (
    SELECT 
        *, 
        CASE WHEN LAG(window) OVER(PARTITION BY category ORDER BY ts  ) = window THEN 1 ELSE 0 END is_group
    FROM @TestTable 
    ) x
ORDER BY x.category,ts

这几乎有效，但对于最后一段，它不会将行号增加超过 2：

【参考方案1】：

这样的？

DECLARE @TestTable TABLE (category INT, ts INT,window int)
INSERT INTO @TestTable (category,ts,window)
VALUES  (1,1,1),(1,2,1),(1,3,0),(1,4,0),(1,5,1),(1,6,1),(1,7,1),(2,1,0),(2,2,1),(2,3,1),(2,4,1),(2,5,0),(2,6,0),(2,7,1),(2,8,1),(2,9,1),(2,10,1),(2,11,1);


with ctex
as
(
SELECT
x.category,
ts,
window, 
is_group,
sum(is_group) over (partition by category order by ts) as groupstot


FROM
(
SELECT 
    *, 
    CASE WHEN LAG(window) OVER(PARTITION BY category ORDER BY ts  ) = window THEN 0 ELSE 1 END  is_group
FROM @TestTable 
) x

)
Select * , row_number() over(partition by category,groupstot order by ts) from ctex
ORDER BY category,ts

【讨论】：

完美，做到了。谢谢