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

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

First, we discount any row that has a value of 10 or less:

WITH t10 AS (SELECT * FROM t WHERE data > 10),

Next, get the rows whose immediate predecessor is also more than 10:

okleft AS (SELECT t10.*, pred.id AS predid FROM
   t10
   INNER JOIN t pred ON 
        pred.[when] < t10.[when]
        AND pred.[when] >= ALL (SELECT [when] FROM t t2 WHERE t2.[when] < t10.[when])
   WHERE pred.data > 10
),

Also get the rows whose immediate successor is also more than 10:

okright as (SELECT t10.*, succ.id AS succid FROM
   t10
   INNER JOIN t succ ON
        succ.[when] > t10.[when] 
        AND succ.[when] <= ALL (SELECT [when] FROM t t2 WHERE t2.[when] > t10.[when])
   WHERE succ.data > 10
),

Finally, select any row where it either starts a sequence of 3, is in the middle of one, or ends one:

A row whose valid right side also has a valid right side starts a sequence of at least 3:

starts3 AS (SELECT id, [when], data FROM okright r1 WHERE EXISTS(
SELECT NULL FROM okright r2 WHERE r2.id = r1.succid)),

A row whose predecessor and successor are both valid is in the middle of at least 3:

mid3 AS (SELECT id, [when], data FROM okleft l WHERE EXISTS(
SELECT NULL FROM okright r WHERE r.id = l.id)),

A row whose valid left side also has a valid left side ends a sequence of at least 3:

ends3 AS (SELECT id, [when], data FROM okleft l1 WHERE EXISTS(
SELECT NULL FROM okleft l2 WHERE l2.id = l1.predid))

Join them all up, with UNION to remove duplicates:

SELECT * FROM starts3
UNION SELECT * FROM mid3
UNION SELECT * FROM ends3

SQL Fiddler: http://sqlfiddle.com/#!3/12f3a/9

Edit: I like BVR's answer, much more elegant than mine.

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

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