I want to display the lowest earning employees of each department based on salary using min().
I have tables of employees with id, first name, last name, department id, salary
and departments, department_id, name department id from 1 to 5.
I am having trouble doing this, I only know how to start
SELECT name, surname from employees WHERE...
You would use min() for this. You would use window functions:
select e.*
from (select e.*,
rank() over (partition by department_id order by salary) as seqnum
from employees e
) e
where seqnum = 1
order by department_id;
With rank()over() ranking window function you can have your rows within a group ranked as you wish. Here we have ranked all the employees starting with lowest salary in a department. Now if we select rows with rn =1 then it will select employees from a department with lowest salary.
Then joined the result with department table to get the name of the
With lowestEarningEmployees as
(
id, first_name, last_name, department_id, salary, rank()over(partition by department_id order by salary)rn from employees
)
select le.id, le.first_name, le.last_name, le.department_id,d.name, le.salary,
from lowestEarningEmployees le inner join departments d on le.department_id=d.department_id
** If more than one employee in a department have lowest salary all of them will be selected. If you want to select only one employee with lowest salary then you need to use row_number() instead of rank().
You can also use subquery to have your desired result (Though I would suggest to use first one) :
Select e.id, e.first_name, e.last_name, e.department_id, d.name, e.salary
from employees e inner join department d on e.department_id = d.department_id
where e.salary=(select min(salary) from employees empl where e.department_id=empl.department_id)
Having two tables
Employee
Id
Name
Salary
DepartmentId
and
Departament
Id
Name
How can I get the highest average salary within two tables
like
Joe and Max belong to dept 1 so, avg is (70K+90K)/2
= 80K
and
Henry and Sam belog to dept 2, avg is (80K + 60K)/2=70k
so How to select the greatest avg salary by depto?, in this case
IT 80K
i have been trying:
'group the salary by each department and use the Max function to obtain the highest one.
select
Department.Name as Department,
T.M as Salary
from
Employee,
Department,
(select DepartmentId as ID, Max(Salary) as M from Employee group by DepartmentId) as T
where
Employee.Salary = T.M and
Department.Id = T.ID and
Employee.DepartmentId = Department.Id
enter image description here
If multiple department having same maximum avg salary then this solution will return multiple rows.
SELECT *
FROM(
SELECT d.Id, d.Name, AVG(e.Salary) avg_salary, RANK() OVER(ORDER BY AVG(e.Salary) DESC) AS rank_
FROM Employee e
INNER JOIN Departament d ON e.DepartmentId = d.Id
GROUP BY d.Id, d.Name
)T
WHERE rank_ = 1
If you want to get the average just for the department, you can use in this way.
select DepartmentId as ID, de.name as Deptname, Avg(Salary) as M from Employee em1
join Department de on de.departmentID = em1.DepartmentId
group by DepartmentId, de.name
If you want employee name along with highest average then you can use this approach as well.
select
Deptname as Department,
e.Name as Employeename,
z.M as Salary
from
Employee e
join
( select DepartmentId,Deptname, M, row_number() (order by m desc) rownum from ( select DepartmentId as ID, de.name as Deptname, Avg(Salary) as M from Employee em1
join Department de on de.departmentID = em1.DepartmentId
group by DepartmentId, de.name) as T) z
on
e.DepartmentId = T.DepartmentId and z.rownum = 1
If you want a full answer, you should provide DDL, sample data and desired result.
If I understand you correctly, you are looking for something like:
SELECT DepartmentID, AVG(Salary) AS AverageSalaryForDept
FROM Employee
GROUP BY DepartmentID
ORDER BY AverageSalaryForDept DESC;
This will give you all the averages, ordered from the highest to the lowest. Now if you want just the top one, add a FETCH clause:
SELECT DepartmentID, AVG(Salary) AS AverageSalaryForDept
FROM Employee
GROUP BY DepartmentID
ORDER BY AverageSalaryForDept DESC
OFFSET 0 ROWS FETCH NEXT 1 ROW ONLY;
HTH
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: