Question

我正在尝试找到一种有效的方法，仅从Expected和Id派生下面的列State。我想要的是，每次Expected为State时，0的数量会增加Id（按+----+-------+----------+ | Id | State | Expected | +----+-------+----------+ | 1 | 0 | 1 | | 2 | 1 | 1 | | 3 | 0 | 2 | | 4 | 1 | 2 | | 5 | 4 | 2 | | 6 | 2 | 2 | | 7 | 3 | 2 | | 8 | 0 | 3 | | 9 | 5 | 3 | | 10 | 3 | 3 | | 11 | 1 | 3 | +----+-------+----------+排序）。

WITH Groups AS 
(
   SELECT Id, ROW_NUMBER() OVER (ORDER BY Id) AS GroupId FROM tblState WHERE State=0
)
SELECT S.Id, S.[State], S.Expected, G.GroupId FROM tblState S
   OUTER APPLY (SELECT TOP 1 GroupId FROM Groups WHERE Groups.Id <= S.Id ORDER BY Id DESC) G

我已经设法通过以下SQL实现了这一点，但是当数据集很大时执行时间很短：

vis_util

是否有更简单，更有效的方法来产生这种结果？（在SQL Server 2012或更高版本中）

Answer 1

只需使用累计金额：

select s.*,
       sum(case when state = 0 then 1 else 0 end) over (order by id) as expected
from tblState s;

Answer 2

其他方法使用subquery：

select *,
       (select count(*)
        from table t1
        where t1.id < t.id and state = 0
       ) as expected
from table t;

当列匹配值时，有效的ROW_NUMBER增量

2 个答案: