select max value in each 5 rows - sql-server

Question

My total rows are variable and not fixed , So there are N rows and I want to separate each 5 rows as a group and select the max value of price in following table in SQL.
Date Price
20170101 100
20170102 110
20170103 90
20170105 80
20170109 76
20170110 50
20170111 55
20170113 80
20170115 100
20170120 99
20170121 88
20170122 98
20170123 120
So in first 5 group the max price is 110 , and second group is 100, and last group max price is 120.

Answer 1

Use a common table expression to group them.
WITH CTE AS (SELECT RANK() OVER (ORDER BY Date) AS Rank, Price
FROM yourtable)
SELECT (Rank - 1) / 5 AS GroupedDate, MAX(Price) AS MAXPRICE
FROM CTE
GROUP BY ((Rank - 1) / 5);
Output
GroupedDate MAXPRICE
0 110
1 100
2 120
SQL Fiddle: http://sqlfiddle.com/#!6/b5857/3/0

Answer 2

You can use row_number as below
;With cte as (
Select *, Bucket = Sum(RowN) over(Order by [date]) from (
Select *, RowN = case when row_number() over(order by [date]) % 5 = 0 then 1 else 0 end from #data1
) a
) Select top (1) with ties [Date], [Price]
from cte
order by row_number() over (partition by Bucket order by Price desc)

Answer 3

You could use:
SELECT grp, MAX(Price) AS price
FROM (SELECT *, ROW_NUMBER() OVER(ORDER BY DATE) / 5 AS grp FROM tab) sub
GROUP BY grp;
-- OUTPUT
grp price
0 110
1 100
2 120
Rextester Demo
*assuming that date is unique
EDIT:
As in something like : 20170101 - 20170109 110
SELECT
CONVERT(VARCHAR(8),MIN(DATE),112) + '-' + CONVERT(VARCHAR(8),MAX(date),112)
, MAX(Price) AS price
FROM (SELECT *, (ROW_NUMBER() OVER(ORDER BY DATE) ) / 5 AS grp FROM tab) sub
GROUP BY grp;
Output:
20170101-20170105 110
20170109-20170115 100
20170120-20170123 120
Rextester Demo2