基于列序列的差距和岛屿查询/重置行计数

Question

我知道有很多这样的例子，但仍然无法找到确切的解决方案。

我正在寻找重置基于序列1和0的行数。

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) 

在上面的表中，我想的行数列由1递增，按类别划分，但是计数需要每次重置窗口的变化。

最好我到目前为止得到的是：

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 

这近作品，但最后一段也不会增加行数进一步比2：

Answer 1

是这样的吗？

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