Find the employee with max salary in his department - sql

Question

Table structure is as below
id dept salary
1 30 2000
2 20 5500
3 30 6700
4 30 8900
5 30 9900
6 10 1120
7 20 8900
8 10 2400
9 30 2600
10 10 2999
I need the output to have two columns:
Id and Salary
Id should be unique and salary should have the max salary

Answer 1

Although data structure is a bit strange, the query below will give the desired result:
create table #tmp(
id numeric(10,0),
dept int,
salary numeric(10,0))
insert #tmp select 1, 30, 2000
insert #tmp select 2, 20, 5500
insert #tmp select 3, 30, 6700
insert #tmp select 4, 30, 8900
insert #tmp select 5, 30, 9900
insert #tmp select 6, 10, 1120
insert #tmp select 7, 20, 8900
insert #tmp select 8, 10, 2400
insert #tmp select 9, 30, 2600
insert #tmp select 10, 10, 2999
select * from #tmp order by dept
id dept salary
------------ ----------- ------------
6 10 1120
8 10 2400
10 10 2999
2 20 5500
7 20 8900
1 30 2000
9 30 2600
3 30 6700
4 30 8900
5 30 9900
select id, salary
from #tmp t1, (select dept=dept, salaryMax=max(salary) from #tmp group by dept) t2
where t1.dept = t2.dept
and t1.salary = t2.salaryMax
id salary
------------ ------------
7 8900
10 2999
5 9900

Answer 2

SELECT ID,Salary FROM TABLE_NAME WHERE SALARY =(SELECT MAX(SALARY) FROM TABLE_NAME)

Answer 3

SELECT a.Id, a.Salary
FROM table a
WHERE a.Salary = (SELECT MAX(Salary) FROM table
WHERE Id = a.Id)

Answer 4

select * from (select * from dep order by salary desc, dep, id) t
group by t.dep

Answer 5

Use NOT EXISTS to return an epmloyee if no other within same department has a higher salary.
select * from table t1
where not exists (select 1 from table t2
where t2.dept = t1.dept
and t2.salary > t1.salary)
Core SQL-99, will work with both Oracle and Sybase!
Alternative solution:
select * from table
where (dept, salary) = (select dept, max(salary)
from table
group by dept)
Using the following feature outside Core SQL-2003: F641, "Row and table constructors". (Don't know what Sybase and Oracle supports here...)

Answer 6

Using single scan:
WITH SALARIES AS (
SELECT 1 ID, 30 DEPT, 2000 SALARY FROM DUAL UNION ALL
SELECT 2 ID, 20 DEPT, 5500 SALARY FROM DUAL UNION ALL
SELECT 3 ID, 30 DEPT, 6700 SALARY FROM DUAL UNION ALL
SELECT 4 ID, 30 DEPT, 8900 SALARY FROM DUAL UNION ALL
SELECT 5 ID, 30 DEPT, 9900 SALARY FROM DUAL UNION ALL
SELECT 6 ID, 10 DEPT, 1120 SALARY FROM DUAL UNION ALL
SELECT 7 ID, 20 DEPT, 8900 SALARY FROM DUAL UNION ALL
SELECT 8 ID, 10 DEPT, 2400 SALARY FROM DUAL UNION ALL
SELECT 9 ID, 30 DEPT, 2600 SALARY FROM DUAL UNION ALL
SELECT 10 ID, 10 DEPT, 2999 SALARY FROM DUAL
)
SELECT
MAX(ID) KEEP (DENSE_RANK LAST ORDER BY SALARY) ID,
MAX(SALARY) SALARY
FROM
SALARIES
GROUP BY
DEPT;

Answer 7

All the below three queries worked :
select t.dept, t.salary
from
(select dept,max(salary) as m_sal from tempdb..test
group by dept)x, tempdb..test t
where x.dept = t.dept
and x.m_sal = t.salary
select dept, id, salary as m_sal
from tempdb..test t1
where salary = (select max(salary) from tempdb..test t2 where t2.dept = t1.dept)
select t1.dept,t1.salary from tempdb..test t1
inner join
(select dept,max(salary) as m_sal from tempdb..test
group by dept)x on
t1.salary=x.m_sal and
t1.dept=x.dept