TSQL - Get Data grouped by row number - sql

Question

I have a table with ID and Date field
ID |Date
1 |2013-5-22
1 |2013-5-23
1 |2013-5-25
1 |2013-5-26
2 |2013-5-26
2 |2013-5-27
1 |2013-5-27
1 |2013-5-28
With the Row_Number i can group all data by id and ghet the Min date and Max Date
;WITH q AS(
SELECT f.*,
grp = DATEDIFF(day, 0, f.Date) - ROW_NUMBER() OVER (PARTITION BY f.ID ORDER BY f.Date),
FROM myTable f
)
SELECT
MIN(q.ID) as ID,
MIN(q.Date) as StartDate,
MAX(q.Date) as EndDate
FROM q
GROUP BY q.grp, q.ID, Date
;
Result:
ID |StartDate |EndDate
1 |2013-5-22 |2013-5-23
2 |2013-5-26 |2013-5-27
1 |2013-5-25 |2013-5-28
Now i need to get the date step by <= 3
Example:
ID |StartDate |EndDate
1 |2013-5-22 |2013-5-23
2 |2013-5-26 |2013-5-27
1 |2013-5-25 |2013-5-27
1 |2013-5-28 |2013-5-28
Can someone, please, illuminate my way?
ty
EDIT
Sorry
;WITH q AS(
SELECT f.*,
grp = DATEDIFF(day, 0, f.Date) - ROW_NUMBER() OVER (PARTITION BY f.ID ORDER BY f.Date)
FROM MyTable f
)
SELECT
MIN(q.ID) as ID,
MIN(q.Date) as StartDate,
MAX(q.Date) as EndDate
FROM q
GROUP BY q.grp, q.ID
;

Answer 1

My first attempt had a bug, try this instead:
;WITH q AS(
SELECT ID, Date,
grp = DATEDIFF(day, 0, Date) - ROW_NUMBER() OVER (PARTITION BY ID ORDER BY Date)
FROM myTable
), r as
(
select id, date, grp,
(ROW_NUMBER() OVER (PARTITION BY grp ORDER BY Date)-1)/3 a from q
)
SELECT
MIN(ID) as ID,
MIN(Date) as StartDate,
MAX(Date) as EndDate
FROM r
GROUP BY grp, ID, a