second highest salary in each department - sql-server

Question

I am trying to find the second highest salary in each department.
Schema:
CREATE TABLE employees
(
ID int NOT NULL,
NAME char(50) NOT NULL,
departmentid int,
salary int
);
Sample records:
/*departmentid =1 */
INSERT INTO employees VALUES (1, 'Max', 1, 90000);
INSERT INTO employees VALUES (2, 'Joe', 1, 70000);
INSERT INTO employees VALUES (3, 'Randy', 1, 70000);
/*departmentid =2 */
INSERT INTO employees VALUES (4, 'Henry', 2, 80000);
INSERT INTO employees VALUES (5, 'SAM', 2, 60000);
/*departmentid =3 */
INSERT INTO employees VALUES (6, 'Janet', 3, 69000);
My query:
SELECT departmentid,
NAME,
salary
FROM
(
SELECT
departmentid,
NAME,
salary,
Dense_rank()OVER (partition BY departmentid
ORDER BY salary DESC) AS Rank,
Count(1)OVER(partition BY departmentid) AS cnt
FROM
employees
)t
WHERE
t.rank = 2
OR ( t.rank = 1
AND cnt = 1 )
The output I am getting is as below;
departmentid NAME salary
1 Joe 70000
1 Randy 70000
2 SAM 60000
3 Janet 69000
My expected output is
departmentid NAME salary
1 Joe 70000
1 Randy 70000
2 SAM 60000
3 NULL NULL
As there is only one record for departmentid=3, it should return null.
What is wrong with this query? Any other ways to achieve this result?
I've also included a SQL fiddle.

Answer 1

ROW_NUMBER() and select = 2
;WITH salary AS
(
[RN] = SELECT ROW_NUMBER() OVER (PARTITION BY departmentid ORDER BY salary),*
FROM <table>
)
SELECT
*
FROM salary
WHERE [RN] = 2

Answer 2

I've used two CTEs.
The first returns a list of every department. You'll need this to ensure departments with less than 2 salaries are included in the final result.
The second ranks each employee within their department.
Finally, I've used a left outer join to maintain the complete list of departments.
WITH Department AS
(
-- Returns a list of the departments.
SELECT
departmentid
FROM
employees
GROUP BY
departmentid
),
EmployeeRanked AS
(
SELECT
DENSE_RANK() OVER (PARTITION BY departmentid ORDER BY salary DESC) AS [Rank],
departmentid,
NAME,
salary
FROM
employees
)
SELECT
er.Rank,
d.departmentid,
er.NAME,
er.salary
FROM
Department AS d
LEFT OUTER JOIN EmployeeRanked AS er ON er.departmentid = d.departmentid
AND er.[Rank] = 2
;
Returns
Rank departmentid NAME salary
2 1 Joe 70000
2 1 Randy 70000
2 2 SAM 60000
(null) 3 (null) (null)

Answer 3

Use a sub query as i wrote here : http://sqlfiddle.com/#!6/bb5e1/26
with ranks as(
SELECT departmentid,
salary,
row_number() over (partition by (departmentid) order by salary desc) as rank
FROM employees
)
Select *
from ranks
Where ranks.rank = 2

Answer 4

If the departmentid having only one row, and if you consider that also. Then
Query
;with cte as(
select [rank] = dense_rank() over(
partition by departmentid
order by departmentid, salary desc
), *
from employees
)
select ID, NAME, departmentid, salary from cte
where [rank] = 2
union all
select max(ID), max(NAME), departmentid, max(salary)
from cte
group by departmentid
having count([rank]) = 1;

Answer 5

There is also a simple way:
SELECT TOP 1 * FROM (Select top 2 * FROM employees order by salary desc ) e Order by salary asc
Edit: this returns only the 2nd highest overall

Answer 6

I think you can get correct answer by just removing below code from your code
OR ( t.rank = 1
AND cnt = 1 )
also main table should be left join from this result to get null in rest of columns