I have a table EmpDetails:
DeptID EmpName Salary
Engg Sam 1000
Engg Smith 2000
HR Denis 1500
HR Danny 3000
IT David 2000
IT John 3000
I need to make a query that find the highest salary for each department.
SELECT DeptID, MAX(Salary) FROM EmpDetails GROUP BY DeptID
The above query is the accepted answer but it will not work for the following scenario. Let's say we have to find the employees with the highest salary in each department for the below table.
DeptID
EmpName
Salary
Engg
Sam
1000
Engg
Smith
2000
Engg
Tom
2000
HR
Denis
1500
HR
Danny
3000
IT
David
2000
IT
John
3000
Notice that Smith and Tom belong to the Engg department and both have the same salary, which is the highest in the Engg department. Hence the query "SELECT DeptID, MAX(Salary) FROM EmpDetails GROUP BY DeptID" will not work since MAX() returns a single value. The below query will work.
SELECT DeptID, EmpName, Salary FROM EmpDetails
WHERE (DeptID,Salary) IN (SELECT DeptID, MAX(Salary) FROM EmpDetails GROUP BY DeptID)
Output will be
DeptID
EmpName
Salary
Engg
Smith
2000
Engg
Tom
2000
HR
Danny
3000
IT
John
3000
Assuming SQL Server 2005+
WITH cteRowNum AS (
SELECT DeptID, EmpName, Salary,
DENSE_RANK() OVER(PARTITION BY DeptID ORDER BY Salary DESC) AS RowNum
FROM EmpDetails
)
SELECT DeptID, EmpName, Salary
FROM cteRowNum
WHERE RowNum = 1;
If you want to show other parameters too along with DeptId and Salary like EmpName, EmpId
SELECT
EmpID
, Name,
, Salary
, DeptId
FROM Employee
where
(DeptId,Salary)
in
(select DeptId, max(salary) from Employee group by DeptId)
SELECT empName,empDept,EmpSalary
FROM Employee
WHERE empSalary IN
(SELECT max(empSalary) AS salary
From Employee
GROUP BY EmpDept)
Select empname,empid,Sal,DeptName from
(Select e.empname,e.empid,Max(S.Salary) Sal,D.DeptName, ROW_NUMBER() Over(partition by D.DeptName order by s.salary desc) Rownum
from emp e inner join Sal S
on e.empid=s.empid
inner join Dept d on e.Deptid=d.Deptid
group by e.empname,e.empid,D.DeptName,s.Salary
) x where Rownum = 1
This will work if the department, salary and employee name are in the same table.
select ed.emp_name, ed.salary, ed.dept from
(select max(salary) maxSal, dept from emp_dept group by dept) maxsaldept
inner join emp_dept ed
on ed.dept = maxsaldept.dept and ed.salary = maxsaldept.maxSal
Is there any better solution than this?
ermn, something like:
select
d.DeptID,
max(e.Salary)
from
department d
inner join employees e on d.DeptID = e.DeptID
group by
d.DeptID
WITH cteRowNum AS (
SELECT DeptID, EmpName, Salary,
ROW_NUMBER() OVER(PARTITION BY DeptID ORDER BY Salary DESC) AS RowNum
FROM EmpDetails
)
SELECT DeptID, EmpName, Salary,Rownum
FROM cteRowNum
WHERE RowNum in(1,2);
SELECT Employee_ID
, First_name
, last_name
, department_id
, Salary
FROM (SELECT Employee_ID
, First_name
, last_name
, department_id
, Salary
, MAX(salary) OVER (PARTITION BY department_id) dept_max_sal
FROM EMPLOYEES) AS Emp
WHERE salary = dept_max_sal;
Use following command;
SELECT A.*
FROM #EmpDetails A
INNER JOIN ( SELECT DeptID ,
MAX(salary) AS salary
FROM #EmpDetails
GROUP BY DeptID
) B ON A.DeptID = B.DeptID
AND A.salary = B.salary
ORDER BY A.DeptID
SELECT DeptID, MAX(Salary)
FROM EmpDetails
GROUP BY DeptID
This query will work fine, but the moment if you want to fetch some others details related to the employee having the highest salary will contradict.
You can use :
SELECT DepatID, a , b, c
FROM EmpDetails
WHERE Salary IN (
SELECT max(Salary)
FROM EmpDetails
GROUP BY DeptID
);
if you will use the previous query it will only reflects the records of the min val except the salary as you have used the max function.
SELECT
DeptID,
Salary
FROM
EmpDetails
GROUP BY
DeptID
ORDER BY
Salary desc
***
> /*highest salary by each dept*/
***
select d.Dept_Name,max(e.salary)
from emp_details as e join Dept_Details as d
on e.d_id=d.Dept_Id
group by d.Dept_Name
select distinct e.d_id,d.Dept_Name
from emp_details as e join Dept_Details as d
on e.d_id=d.Dept_Id
select e.salary,d.Dept_Name,d.Dept_Id
from emp_details as e join Dept_Details as d
on e.d_id=d.Dept_Id
/////simplest query for max salary dept_wise////
Use the below quesry:
select employee_name,salary,department_id from emp where salary in(select max(salary) from emp group by department_id);
select empno
from EMP e
where salary=(select max(sal)
from EMP w
where groupby w.deptno having e.deptno=w.deptno)
I hope it will work...
Use correlated subquery:
SELECT DeptID, EmpName, Salary
FROM EmpDetails a
WHERE Salary = (SELECT MAX(Salary)
FROM EmpDetails b
WHERE a.DeptID = b.DeptID)
This is the best possible solution for ORACLE:
Select * from (select customerid, city, freight,
row_number() over (partition by customerid order by freight desc) Row_Number from
(select orders.orderId, customers.CUSTOMERID, customers.city, orders.FREIGHT from orders inner join customers on orders.customerid = customers.customerid where customers.country='Germany' order by customers.customerid, orders.freight desc)
order by customerid, freight desc) where Row_Number<=2;
Notice here I have used partition by clause for marking row number, this is majorly because we need to partition the records grouping them according to customer id. I have used two inner queries here. The inner most query is to give a view which is sorted according to customer ID and decreasing order of cost. Now from that we need to obtain always top two records so firstly we need to name them and then we need to filter them according to rownum. Second level query is to mark rownum according to customer ID. And final query will filter the result according to rownum. For every partition.
select deptid, empname, salary from
(Select deptid, empname,salary,
rank() Over(Partition by deptid order by salary desc)as rank from
EmpDetails) emp
where emp.rank = 1
First ranks each employee by salary in descending order having highest
rank 1 and then selects only deptid, empname, salary. You can do this for
all Nth member of the group.
SELECT empname
FROM empdetails
WHERE salary IN(SELECT deptid max(salary) AS salary
FROM empdetails
group by deptid)
select a.*
from EmpDetails a
inner join
(
select DeptID,max(Salary) as Salary
from EmpDetails group by DeptID
)b
on a.DeptID = b.DeptID and a.Salary = b.Salary
Here is a way to get maximum values and names on any version of SQL.
Test Data:
CREATE TABLE EmpDetails(DeptID VARCHAR(10), EmpName VARCHAR(10), Salary DECIMAL(8,2))
INSERT INTO EmpDetails VALUES('Engg','Sam',1000)
INSERT INTO EmpDetails VALUES('Engg','Smith',2000)
INSERT INTO EmpDetails VALUES('HR','Denis',1500)
INSERT INTO EmpDetails VALUES('HR','Danny',3000)
INSERT INTO EmpDetails VALUES('IT','David',2000)
INSERT INTO EmpDetails VALUES('IT','John',3000)
Example:
SELECT ed.DeptID
,ed.EmpName
,ed.Salary
FROM (SELECT DeptID, MAX(Salary) MaxSal
FROM EmpDetails
GROUP BY DeptID)AS empmaxsal
INNER JOIN EmpDetails ed
ON empmaxsal.DeptID = ed.DeptID
AND empmaxsal.MaxSal = ed.Salary
Not the most elegant, but it works.
SELECT D.DeptID, E.EmpName, E.Salary
FROM Employee E
INNER JOIN Department D ON D.DeptId = E.DeptId
WHERE E.Salary IN (SELECT MAX(Salary) FROM Employee);
select * from (
select a.* from EmpDetails a
right join (select DeptID,max(salary) as Salary from EmpDetails group by DeptID) b
on b.DeptID=a.DeptID and b.salary=a.salary ) as c group by c.DeptID;
The below query will display employee name with their respective department name in which that particular employee name is having highest salary.
with T as
(select empname, employee.deptno, salary
from employee
where salary in (select max(salary)
from employee
group by deptno))
select empname, deptname, salary
from T, department
where T.deptno=department.deptno;
I executed the above query successfully on Oracle database.
If you just want to get the highest salary from that table, by department:
SELECT MAX(Salary) FROM TableName GROUP BY DeptID
IF you want Department and highest salary, use
SELECT DeptID, MAX(Salary) FROM EmpDetails GROUP BY DeptID
if you want more columns in employee and department, use
select Department.Name , emp.Name, emp.Salary from Employee emp
inner join (select DeptID, max(salary) [salary] from employee group by DeptID) b
on emp.DeptID = b.DeptID and b.salary = emp.Salary
inner join Department on emp.DeptID = Department.id
order by Department.Name
if use salary in (select max(salary...)) like this, one person have same salary in another department then it will fail.
The below listed query will list highest salary in each department.
select deptname, max(salary) from department, employee where
department.deptno=employee.deptno group by deptname;
I executed this query successfully on Oracle database.
with ctesal as (
select DepartmentId , Name , Salary, ROW_Number() OVER (partition by DepartmentId
order by Salary desc) as RowNum
from dbo.Employee
)
select DepartmentId , Name , Salary , RowNum from ctesal where RowNum =2;
This is applicable to SQL server.
ROW_Number is a inbuilt function in SQL server .It gives count starting from 1 based on partition by and order by clause. At the end, We can write where condition based on our requirements.
I have like 2 approaches using one with Rank and the other with ROW_NUMBER
This is my sample data
Age Name Gender Salary
----------- -------------------------------------------------- ---------- -----------
1 Mark Male 5000
2 John Male 4500
3 Pavan Male 5000
4 Pam Female 5500
5 Sara Female 4000
6 Aradhya Female 3500
7 Tom Male 5500
8 Mary Female 5000
9 Ben Male 6500
10 Jodi Female 7000
11 Tom Male 5500
12 Ron Male 5000
13 Ramani Female 7000
So here is my first query to find max salary and the person with that max salary for each Gender
with CTE as(
select RANK() over(partition by Gender Order by Salary desc) as [Rank],* from employees)
select * from CTE where [Rank]=1
Rank Age Name Gender Salary
-------------------- ----------- -------------------------------------------------- ---------- -----------
1 10 Jodi Female 7000
1 13 Ramani Female 7000
1 9 Ben Male 6500
So in this case, we can see there is a tie between these 2 female employees "Jodi" and "Ramani". In that case, As a tie-breaker I want to make use of Age as a deciding factor and person with more age is supposed to be displayed
with CTE as(
select RANK() over(partition by Gender Order by Salary desc,age desc) as [Rank],* from employees)
select * from CTE where [Rank]=1
Rank Age Name Gender Salary
-------------------- ----------- -------------------------------------------------- ---------- -----------
1 13 Ramani Female 7000
1 9 Ben Male 6500
Usually, in this case for finding the highest salary, it doesn't make much difference even if
Rank, Dense_Rank, or Row_Number() are used. But they have some impact in other cases.
Thank you #JoeStefanelli for his answer (https://stackoverflow.com/a/8477083/4691279). He provided SQL Server 2005+ version and I used the same to create the Oracle version:
WITH cteRowNum(dep_id, emp_id, Salary, RowNums) AS (
SELECT dep_id, emp_id, Salary,
DENSE_RANK() OVER(PARTITION BY dep_id ORDER BY Salary DESC) AS RowNums
FROM employee
)
SELECT cteRowNum.dep_id, cteRowNum.emp_id, cteRowNum.Salary
FROM cteRowNum
WHERE cteRowNum.RowNums = 1;
You can test this using livesql.oracle.com, below are my DDLs and DMLs you can use:
create table employee (
emp_id varchar2(50) NOT NULL,
dep_id varchar2(50) NOT NULL,
salary number not null
);
create table department (
dep_id varchar2(50) NOT NULL,
dep_name varchar2(50) NOT NULL
);
insert into employee values (100, 5000, 1000000);
insert into employee values (200, 5000, 2000000);
insert into employee values (300, 5000, 3000000);
insert into employee values (400, 6000, 1500000);
insert into employee values (500, 6000, 1500000);
insert into employee values (600, 7000, 1000000);
insert into employee values (700, 7000, 1000000);
insert into employee values (800, 7000, 2000000);
insert into department values (5000, 'dep 1');
insert into department values (6000, 'dep 2');
insert into department values (7000, 'dep 3');
And below is the success screenshot of the query:
This question already has answers here:
How to find the employee with the second highest salary?
(5 answers)
Closed 2 years ago.
Well it is a well known question. Consider the below
EmployeeID EmployeeName Department Salary
----------- --------------- --------------- ---------
1 T Cook Finance 40000.00
2 D Michael Finance 25000.00
3 A Smith Finance 25000.00
4 D Adams Finance 15000.00
5 M Williams IT 80000.00
6 D Jones IT 40000.00
7 J Miller IT 50000.00
8 L Lewis IT 50000.00
9 A Anderson Back-Office 25000.00
10 S Martin Back-Office 15000.00
11 J Garcia Back-Office 15000.00
12 T Clerk Back-Office 10000.00
We need to find out the second highest salary
With Cte As
(
Select
level
,Department
,Max(Salary)
From plc2_employees
Where level = 2
Connect By Prior (Salary) > Salary)
Group By level,Department
)
Select
Employeeid
,EmployeeName
,Department
,Salary
From plc2_employees e1
Inner Join Cte e2 On e1.Department = e2.Department
Order By
e1.Department
, e1.Salary desc
,e1.EmployeeID
is somehow not working... I am not getting the correct result. Could anyone please help me out.
Something like
select * from
(
select EmployeeID, EmployeeName, Department, Salary,
rank () over (partition by Department order by Salary desc) r
from PLC2_Employees
)
where r = 2
Edit - tested it and it gives the answer you expected.
If you're going to teach yourself how to deal with CONNECT BY, you should first find a problem that is suited to the construct. CONNECT BY is meant for processing data that's in a hierarchical form, which your example is not. Salaries are not related to each other in a hierarchical fashion. Trying to force-fit a construct on the wrong problem is frustrating and doesn't really teach you anything.
Take a look at the classic employee-manager relationship in the demo HR schema you can install with Oracle. All employees report to a manager, including managers (except the top guy). You can then use this schema to create a query to show, for example, the Organization Chart for the company.
START WITH … CONNECT BY is designed to explore data that forms a graph, by exploring all possible descending paths. You specify the root nodes in the START WITH clause and the node connections in the CONNECT BY clause (not in the WHERE clause).
The WHERE clause filters will be processed after the hierachical conditions, same for GROUP BY and HAVING (of course because GROUP BY is computed after WHERE).
Therefore you MUST here CONNECT BY PRIOR department = department for example. You must also avoid that a node connection is done between two salaries when there is an intermediate salary.
Therefore the final query would resemble this:
SELECT level
, Department
, Salary
FROM plc2_employees pe1
START WITH pe1.salary = (select max(salary) from plc2_employees pe2 WHERE pe2.Department = pe1.Department)
CONNECT BY PRIOR pe1.Department = pe1.Department
AND PRIOR pe1.Salary > pe1.Salary
AND PRIOR pe1.Salary = ( SELECT MIN(Salary) FROM plc2_employees pe3
WHERE pe3.Department = pe1.Department
AND pe3.Salary > pe1.Salary
)
The recursion condition states that there is no intermediate salary between the child row and the parent row.
Note that this will really be unefficient…
Try this, it gives second highest salary
select MAX(Salary) as Salary
from Employee_salary
where Salary not in (select MAX(Salary) from Employee_salary)
You can use this query:
select * from
employee e1
where 2 = (select count (distinct (salary))
from employee e2
where e2.salary >=e1.salary);
find out second highest salary from employee table having column as salary:
Database : DB2
with t as
(
select distinct salary from employee order by salary desc
),
tr as
(
select salary, row_Number() over() r from t
)
select salary from tr where r = 2
Try this,
It gives second highest salary...
select MAX(Salary) as Salary
from Employee_salary
where Salary not in (select MAX(Salary) from Employee_salary )
If you want to find nth highest salary than you can use following query....
you need to do just one change.....
Put the value of N=nth highest
Cheers....:)
SELECT * FROM Employee_salary Emp1
WHERE (N-1) = (SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee_salary Emp2
WHERE Emp2.Salary > Emp1.Salary)
This will work -
SELECT MIN(Salary)
FROM employee
WHERE salary IN (SELECT TOP 2 salary FROM employee ORDER BY salary DESC)
First, select the distinct salaries in descending order (from greatest to least), from that set select the top 2 and put in ascending order (placing number 2 on top), then from those 2 select top 1:
select top 1 s.Salary
from
(select top 2 t.Salary
from
(select distinct Salary
from PLC2_Employees
order by Salary desc) t
order by Salary asc) s