Question

在连续ID的测试中构建是很困难的，不会将其分解成部分或使用我想避免的光标。

伪查询 -

LazySeq

期望的结果

SELECT all 
FROM table with the same description on multiple adjacent rows for >= 4 rows 
and set tag = 'y' and order by id 

(id,description, tag),
(1, 'xxx', 'n'),
(2, 'xxx', 'n'),
(3, 'xxx', 'n'),
(7, 'xxx', 'n'),
(5, 'xxx', 'n'),
(8, 'xxx', 'n'), 
(4, 'xxx', 'n'), 
(6, 'zzz', 'n')

Answer 1

这被称为间隙和岛屿问题。这样的事情应该起作用

;with cte as
(SELECT id, 
       description, 
       tag = 'y' ,
       cnt = Count(*)over(partition by description, grp)
FROM  (SELECT *, 
              grp = Sum(CASE WHEN prev_description = description THEN 0 ELSE 1 END)Over(Order by id)
       FROM   (SELECT *, 
                      prev_description = Lag(description) OVER(ORDER BY id) 
               FROM   Yourtable) a) b 
GROUP  BY id, description, grp 
)
Select * from cte 
Where cnt >= 4

使用Row_Number

的另一种方法

;with cte as
(SELECT id, 
       description, 
       tag = 'y' ,
       cnt = Count(*)over(partition by description, grp)
FROM  (select Grp = row_number()over(order by id) - 
             row_number()over(partition by description order by id), *
       from Yourtable) b 
GROUP  BY id, description, grp)
Select * from cte 
Where cnt >= 4

Answer 2

我认为这样做会

select *, 'y' as 'newTag' 
from ( select *
           , count(*) over (partition by [description], grp) as 'grpSize' 
       from ( select * 
                   , ( [id] - row_number() over (partition by [description] order by [id]) ) as grp
              from [consecutive] 
            ) tt
     ) ttt 
where grpSize >= 4
order by [description], grp, [id]

如果id为n行连续，如何标记一组重复项？

2 个答案: