Schema for EMPLOYEE
(ID, EMPLOYEENAME, SALARY, ORGANIZATIONID)
Query to Solve: Find employee Names in each organization with Maximum Salary without a Join.
SELECT E.*
FROM EMPLOYEE E,
(SELECT EMP.ORGANIZATIONID, MAX(EMP.SALARY)
FROM EMPLOYEE EMP
GROUP BY EMP.ORGANIZATIONID) MAXSALARY
WHERE MAXSALARY.SALARY =E.SALARY
AND E.ORGANIZATIONID=EMP.ORGANIZATIONID ;
Is there a way to avoid the join? I am using Spark SQL API and joins cause an extra shuffle operation which is expensive. Is there a way to get the employee name while getting the max salary?
Assume you have a single employee in each organization having the max salary
You can use PARTITION BY with Spark SQL as shown below (Although it will require a subquery)
SELECT E.*
FROM
(SELECT EMP.EMPLOYEENAME, EMP.ORGANIZATIONID, EMP.SALARY,
row_number() OVER (PARTITION BY ORGANIZATIONID ORDER BY SALARY DESC) as rank
FROM EMPLOYEE EMP
) AS E
WHERE E.rank=1
Try this:
SELECT P.ORGANIZATIONID, P.EMPLOYEENAME
FROM EMPLOYEE P
WHERE P.SALARY = (SELECT MAX(E.SALARY) FROM EMPLOYEE E WHERE P.ORGANIZATIONID = E.ORGANIZATIONID)
GROUP BY P.ORGANIZATIONID, P.EMPLOYEENAME
Try this:
SELECT EMPLOYEENAME FROM EMPLOYEE
WHERE SALARY IN (SELECT MAX(SALARY) FROM EMPLOYEE GROUP BY ORGANIZATIONID)
Related
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
A table, Employee has columns EmployeeID, Salary.
How do I find the EmployeeIDs which have a salary greater than the average salary?
Using Subqueries:
SELECT EmployeeID
FROM Employee
WHERE Salary > (SELECT AVG(Salary)
FROM Employee);
Is it possible using joins?
Is any other method possible?
It seems really silly to me to write it this way, but here it is without any subqueries:
SELECT a.EmployeeID
FROM Employee a
CROSS JOIN Employee b
GROUP BY a.EmployeeID, a.Salary
HAVING a.Salary > AVG(b.Salary)
SELECT EmployeeID
FROM Employee, (SELECT AVG(Salary) avg_savary
FROM Employee) sal
WHERE Salary > sal.avg_savary;
Move the average salary calculation to FROM to calculate it once. BTW most of moder DB can optimize your query to calculate it once.
SELECT * from
(SELECT EmployeeID,salary AVG(salary) OVER() avg_salary FROM Employee)
WHERE Salary >avg_salary
SELECT e.EmployeeID FROM Employee e
JOIN
(
SELECT avg(Salary) as Salary FROM Employee
) e1
ON e.Salary > e1.Salary
Declare #AverageSalary as Money
Select #AverageSalary = AVG(Salary) From Employee
Select * from Employee Where Salary > #AverageSalary
We have only a table named EMPLOYEESALARY in our database with the 3 following columns:
Employee_ID, Employee_Salary, Department_ID
Now I have to SELECT every employee that has a higher salary than the AVERAGE of his department. How do I do that?
I know this is a repeat question but the best solution I found everywere was:
SELECT * from employee join (SELECT AVG(employee_salary) as sal, department_ID
FROM employee GROUP BY Department_ID) as t1
ON employee.department_ID = t1.department_ID
where employee.employee_salary > t1.sal
Can we optimize it further and do it without a subquery?
Reference:
SELECT every employee that has a higher salary than the AVERAGE of his department
Employees with higher salary than their department average?
Find Schema here, to test: SQL Fiddle
Can we do it without a subquery?
Not that I can think of. Had the condition been >= then the following would have worked
SELECT TOP 1 WITH TIES *
FROM employee
ORDER BY CASE
WHEN employee_salary >= AVG(employee_salary)
OVER (
PARTITION BY Department_ID) THEN 0
ELSE 1
END
But this is not an optimisation and it won't work correctly for the > condition if no employee has a salary greater than the average anyway (i.e. all employees in a department had the same salary)
Can we optimize it further?
You could shorten the syntax a bit with
WITH T AS
(
SELECT *,
AVG(employee_salary) OVER (PARTITION BY Department_ID) AS sal
FROM employee
)
SELECT *
FROM T
WHERE employee_salary > sal
but it still has to do much the same work.
Assuming suitable indexes on the base table already exist then the only way of avoiding some more of that work at SELECT time would be to pre-calculate the grouped SUM and COUNT_BIG in an indexed view grouped by Department_ID (to allow the average to be cheaply derived) .
A more optimal form is likely to be:
select e.*
from (select e.*, avg(employee_salary) over (partition by department_id) as avgs
from employee e
) e
where employee_salary > avgs;
This (as well as other versions) can use an index on employee(department_id, employee_salary). The final where probably should not use an index, because it is selecting lots of rows.
I have two tables
EMPLOYEE (Fname, Lname, Ssn, Salary, Dno)
DEPARTMENT (Dname, Dno, Location)
I want to list the names of all employees making the least in their department
I have come up with this
select min(E.Salary) from EMPLOYEE E group by E.Dno;
but how do I join the EMPLOYEE table with it and display the 'Fname' and 'Lname';
Analytic functions would be best but this would also work:
select *
from employee e
where salary = (select min(x.salary) from employee x where x.dno = e.dno)
Use Analytic function, ROW_NUMBER() OVER( PARTITION BY DNO ORDER BY SALARY) as RN. So, WHERE RN = 1 will give you the employee with least salary in each department.
Remember, if there are two employees with same salary, then you need to use DENSE_RANK to avoid similar rank.
Note : This answer is for Oracle.
A simple way to do it is to check that no one with a lower salary exists in the same department;
SELECT e1.*
FROM employee e1
WHERE NOT EXISTS(
SELECT 1 FROM employee e2 WHERE e1.dno = e2.dno AND e1.salary > e2.salary
);
An SQLfiddle to test with.
How can I get id of department in which employees receive the maximum salary:
Employee table: Empl (ID, FirstName, LastName, Salary, DeptId)
Departments table: Dept (ID, City)
rus (Вывести “id” подразделения, в котором сотрудники получают максимальную заработную плату.)
EDIT: Changed SUM(Salary) to AVG(Salary) based on comments on the question.
SELECT TOP 1 DeptId
FROM Employees
GROUP BY DeptId
ORDER BY AVG(Salary) DESC
SELECT TOP 1 B.*
FROM (SELECT DeptId, AVG(Salary) AvgSalary
FROM Empl
GROUP BY DeptId) A
INNER JOIN Dept B
ON A.DeptId = B.Id
ORDER BY AvgSalary DESC
To get the one single Department's ID where the highest single salary is paid:
SELECT TOP 1 DeptID
FROM dbo.Empl
ORDER BY Salary DESC
Or are you looking for something else?
I would assume you mean max average salary of a department and not the single highest salary across all departments.
However it seems all you would have to do is use the following SQL functions
MAX function
AVG function
group by department ID and viola.
Thought I agree with the comments above, I will assume you are doing this for research ;-)
select id
from dept
where id = ( select deptid
from ( select max(avg_salary), deptid
from ( select deptid, avg(salary) as avg_salary
from empl
group by deptid )
group by deptid )
)
:-)
SELECT DepartmentId
FROM Employee
WHERE Salary = (SELECT MAX(Salary) FROM Employee)