Maximum and Minimum Rows Alternatively in SQL Server - sql-server

Question

This is an Employee table,
Id Name Salary
1 A.J 7000
2 B.S 30000
3 C.K 2000
4 D.O 10000
5 E.L 500
Now i want to display 1st highest salary then minimum salary then 2nd maximum salary then 2nd minimum salaray and so on..up to nth row.
Expected Output,
Id Name Salary
2 B.S 30000
5 E.L 500
4 D.O 10000
3 C.K 2000
1 A.J 7000

Answer 1

One more variant without explicit COUNT. SQL Fiddle.
Try also to add this row to sample data (6, 'X.Y', 7000) in the fiddle. The query still returns correct results.
DECLARE #Employee TABLE (ID int, Name nvarchar(50), Salary money);
INSERT INTO #Employee (ID, Name, Salary) VALUES
(1, 'A.J', 7000),
(2, 'B.S', 30000),
(3, 'C.K', 2000),
(4, 'D.O', 10000),
(5, 'E.L', 500);
WITH
CTE
AS
(
SELECT *, NTILE(2) OVER (ORDER BY Salary, ID) AS n
FROM #Employee AS E
)
SELECT
*
,SIGN(n-1.5) AS s
,SIGN(n-1.5)*Salary AS ss
,ROW_NUMBER() OVER(PARTITION BY n ORDER BY SIGN(n-1.5)*Salary DESC) AS rn
FROM CTE
ORDER BY rn, ss DESC;
Result
ID Name Salary n s ss rn
2 B.S 30000.00 2 1.0 30000.00000 1
5 E.L 500.00 1 -1.0 -500.00000 1
4 D.O 10000.00 2 1.0 10000.00000 2
3 C.K 2000.00 1 -1.0 -2000.00000 2
1 A.J 7000.00 1 -1.0 -7000.00000 3
I left intermediary columns in the output to illustrate how it works.

Answer 2

Using Row_Number() and Count()
Fiddle Demo
declare #count int=(select count(1) from Employee);
with cte1 as
(
select ROW_NUMBER() over(order by salary desc) as rn,0 Sort,Id,Name,Salary, count(Id) over () cnt from Employee
union all
select ROW_NUMBER() over(order by salary) as rn,1 Sort,Id,Name,Salary, count(Id) over () cnt from Employee
)
select top (#count) Id,Name,Salary from cte1 where rn <= (floor(cnt/2) + cnt%2) order by rn,sort

Answer 3

Below is the solution:
--Create dummy employee table
CREATE TABLE tbl_Employee
(
Id INT,
Name VARCHAR(100),
Salary NUMERIC(9, 2)
)
GO
--Insert few dummy rows in the table
INSERT INTO #Employee
(Id, Name, Salary)
VALUES(100, 'John', 7000),
(101, 'Scott', 30000),
(102, 'Jeff', 2000),
(103, 'Jimy', 10000),
(104, 'Andrew', 500),
(105, 'Alister', 100)
GO
--Get data as required
DECLARE #Cnt INT = 0, #SeqLimit INT = 0
SELECT #Cnt = COUNT(1) FROM tbl_employee
SET #SeqLimit = CEILING(#Cnt / 2.0)
SELECT * FROM
(
SELECT ROW_NUMBER() OVER(ORDER BY Salary DESC) AS SEQ, Id, Name, Salary FROM tbl_employee
)DT1
WHERE SEQ <= #SeqLimit
UNION ALL
SELECT * FROM
(
SELECT ROW_NUMBER() OVER(ORDER BY Salary ASC) AS SEQ, Id, Name, Salary FROM tbl_employee
)DT2
WHERE SEQ <= #SeqLimit - (#Cnt % 2)
ORDER BY SEQ ASC, Salary DESC
The same can be achieved with different approaches and here you can find more on this:
http://www.sqlrelease.com/order-max-and-min-value-rows-alternatively-in-sql-server