How to tag a group of repeating items if the ids are consecutive for n rows?

问题

Building in the test for consecutive ids is proving difficult without breaking it down into parts or using a cursor which I'd like to avoid.

pseudo query -

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')

desired result

(1, 'xxx', 'y') 
(2, 'xxx', 'y') 
(3, 'xxx', 'y') 
(4, 'xxx', 'y') 
(5, 'xxx', 'y')

回答1:

This is called as gaps and island problem. Something like this should work

;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

Another approach using Row_Number

;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

回答2:

I think this will do it

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]

来源：https://stackoverflow.com/questions/48357355/how-to-tag-a-group-of-repeating-items-if-the-ids-are-consecutive-for-n-rows

标签

sql

sql-server

gaps-and-islands