How to get the first and the last record per group in SQL Server 2008?

Question

As the title suggests, I\'d like to select the first and last row of each set of rows grouped with a GROUP BY.

I\'ve this table with the following data

Answer 1

select tt.id, tt.groups, tt.val, x.sstart, tt.[end] from test_table tt join
(Select groups,First_value(start) over (partition by groups order by groups) sstart from test_table  
Union
Select groups,Last_Value(start) over (partition by groups order by groups) sstart from test_table) as x
on tt.start=x.sstart
Order by tt.groups, sstart Desc

Answer 2

How to two query 'UNION'

SELECT TOP 1 EmployeeId, AttendenceId, Intime
FROM EmployeeAttendence
WHERE AttendenceDate >='1/18/2020 00:00:00' 
  AND AttendenceDate <='1/18/2020 23:59:59'
GROUP BY EmployeeId,AttendenceId,Intime
ORDER BY AttendenceId  

SELECT TOP 1 EmployeeId, AttendenceId, OutTime
FROM EmployeeAttendence
WHERE AttendenceDate >='1/18/2020 00:00:00' 
  AND AttendenceDate <='1/18/2020 23:59:59'
GROUP BY EmployeeId, AttendenceId, OutTime
ORDER BY AttendenceId desc 

Answer 3

This is one way -

select  t.*
from    tbl t
join    (
        select [group],
               min(val) as val_1,
               max(val) as val_2
        from   tbl
        group by [group]
        ) v
     on t.[group] = v.[group]
    and (t.val = v.val_1
     or t.val = v.val_2);

Fiddle: http://sqlfiddle.com/#!3/c682f/1/0

Another approach:

select id, [group], val, [start], [end]
from(
select t.*,
       max(val) over(partition by [group]) as max_grp,
       min(val) over(partition by [group]) as min_grp
from tbl t
) x
where val in (max_grp,min_grp)

Answer 4

How about using ROW_NUMBER:

SQL Fiddle

WITH Cte AS(
    SELECT *,
        RnAsc = ROW_NUMBER() OVER(PARTITION BY [group] ORDER BY val),
        RnDesc = ROW_NUMBER() OVER(PARTITION BY [group] ORDER BY val DESC)
    FROM tbl
)
SELECT
    id, [group], val, start, [end]
FROM Cte
WHERE
    RnAsc = 1 OR RnDesc = 1
ORDER BY [group], val