sql - How to find consecutive rows based on the value of a column?

Question

I have some data. I want to group them based on the value of data column. If there are 3 or more consecutive rows that have data bigger than 10, then those rows are what I want.

So for this data:

use tempdb;
go
set nocount on;

if object_id('t', 'U') is not null
drop table t;
go

create table t
(
    id int primary key identity,
    [when] datetime,
    data int
)
go

insert into t([when], data) values ('20130801', 1);
insert into t([when], data) values ('20130802', 121);
insert into t([when], data) values ('20130803', 132);
insert into t([when], data) values ('20130804', 15);
insert into t([when], data) values ('20130805', 9);
insert into t([when], data) values ('20130806', 1435);
insert into t([when], data) values ('20130807', 143);
insert into t([when], data) values ('20130808', 18);
insert into t([when], data) values ('20130809', 19);
insert into t([when], data) values ('20130810', 1);
insert into t([when], data) values ('20130811', 1234);
insert into t([when], data) values ('20130812', 124);
insert into t([when], data) values ('20130813', 6);

select * from t;

What I want is:

id          when                    data       
----------- ----------------------- -----------
2           2013-08-02 00:00:00.000 121        
3           2013-08-03 00:00:00.000 132        
4           2013-08-04 00:00:00.000 15         
6           2013-08-06 00:00:00.000 1435       
7           2013-08-07 00:00:00.000 143        
8           2013-08-08 00:00:00.000 18         
9           2013-08-09 00:00:00.000 19    

How to do that?

Answer 1

Try this

WITH cte
AS
(
    SELECT *,COUNT(1) OVER(PARTITION BY cnt) pt  FROM
    (
        SELECT tt.*
           ,(SELECT COUNT(id) FROM t WHERE data <= 10 AND ID < tt.ID) AS cnt
        FROM  t tt
        WHERE data > 10
    ) t1
)

SELECT id, [when], data FROM cte WHERE pt >= 3

SQL FIDDLE DEMO

OUTPUT

id  when                    data
2   2013-08-02 00:00:00.000 121
3   2013-08-03 00:00:00.000 132
4   2013-08-04 00:00:00.000 15
6   2013-08-06 00:00:00.000 1435
7   2013-08-07 00:00:00.000 143
8   2013-08-08 00:00:00.000 18
9   2013-08-09 00:00:00.000 19

EDIT

First the inner query counts the no of records where data <= 10

SELECT tt.*
     ,(SELECT COUNT(id) FROM t WHERE data <= 10 AND ID < tt.ID) AS cnt
FROM  t tt

output

id  when                    data   cnt
1   2013-08-01 00:00:00.000 1       1
2   2013-08-02 00:00:00.000 121     1
3   2013-08-03 00:00:00.000 132     1
4   2013-08-04 00:00:00.000 15      1
5   2013-08-05 00:00:00.000 9       2
6   2013-08-06 00:00:00.000 1435    2
7   2013-08-07 00:00:00.000 143     2
8   2013-08-08 00:00:00.000 18      2
9   2013-08-09 00:00:00.000 19      2
10  2013-08-10 00:00:00.000 1       3
11  2013-08-11 00:00:00.000 1234    3
12  2013-08-12 00:00:00.000 124     3
13  2013-08-13 00:00:00.000 6       4

Then we filter the records with data > 10

WHERE data > 10

Now we count the records by partitoning cnt column

SELECT *,COUNT(1) OVER(PARTITION BY cnt) pt  FROM
(
    SELECT tt.*
        ,(SELECT COUNT(id) FROM t WHERE data <= 10 AND ID < tt.ID) AS cnt
    FROM  t tt
    WHERE data > 10
) t1

Output

id  when    data                   cnt  pt
2   2013-08-02 00:00:00.000 121     1   3
3   2013-08-03 00:00:00.000 132     1   3
4   2013-08-04 00:00:00.000 15      1   3
6   2013-08-06 00:00:00.000 1435    2   4
7   2013-08-07 00:00:00.000 143     2   4
8   2013-08-08 00:00:00.000 18      2   4
9   2013-08-09 00:00:00.000 19      2   4
11  2013-08-11 00:00:00.000 1234    3   2
12  2013-08-12 00:00:00.000 124     3   2

The above query is put in cte just like temp table

Now select the records that are having the consecutive count >= 3

SELECT id, [when], data FROM cte WHERE pt >= 3

ANOTHER SOLUTION

;WITH partitioned AS (
  SELECT *, id - ROW_NUMBER() OVER (ORDER BY id) AS grp
  FROM t
  WHERE data > 10
),
counted AS (
  SELECT *, COUNT(*) OVER (PARTITION BY grp) AS cnt
  FROM partitioned
)

SELECT id, [when], data
FROM counted
WHERE cnt >= 3

Reference URL

SQL FIDDLE DEMO