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)
Related
enter image description here
select Fname,Lname,salary,ssn
from Employee
where joptype='nurse' AND salary <= ALL (select salary
from Employee E,Nurses N
where E.Ssn=N.Ssn AND E.joptype='nurse' AND N.shift='morning')
I want to return a list of nurses name who works at morning and their salary is less than all nurses, or the opposite. I've tried both morning and night, also I've tried greater than all nurses. I think the problem is in the second where it seems like he ignores the shift condition.
You can use EXISTS to find the morning shifts and, from Oracle 12, you can use ORDER BY salary ASC FETCH FIRST ROW WITH TIES to find the matching employees with the lowest salary:
SELECT Fname,
Lname,
salary,
ssn
FROM Employee e
WHERE jobtype='nurse'
AND EXISTS( SELECT 1
FROM Nurses N
WHERE E.Ssn=N.Ssn
AND N.shift='morning' )
ORDER BY salary ASC
FETCH FIRST ROW WITH TIES;
or, if you want to check all nurses and find the minimum salary and then filter on morning shifts:
SELECT *
FROM (
SELECT Fname,
Lname,
salary,
ssn
FROM Employee
WHERE jobtype='nurse'
ORDER BY salary ASC
FETCH FIRST ROW WITH TIES
) e
WHERE EXISTS( SELECT 1
FROM Nurses N
WHERE E.Ssn=N.Ssn
AND N.shift='morning' )
If you want to use a JOIN, rather than EXISTS, then you can use:
SELECT e.Fname,
e.Lname,
e.salary,
e.ssn
FROM (
SELECT Fname,
Lname,
salary,
ssn
FROM Employee
WHERE jobtype='nurse'
ORDER BY salary ASC
FETCH FIRST ROW WITH TIES
) e
INNER JOIN Nurses N
ON E.Ssn=N.Ssn
WHERE N.shift='morning'
(However, if there are multiple entries in Nurses for an Employee then you will get duplicates using a JOIN that you would not get using EXISTS)
Or for your code, you appear to have the filter for shifts in the wrong place:
SELECT e.Fname,
e.Lname,
e.salary,
e.ssn
FROM Employee e
INNER JOIN Nurses n
ON (e.Ssn = n.Ssn)
WHERE e.jobtype='nurse'
AND n.shift='morning'
AND e.salary <= ALL ( SELECT salary
FROM Employee
WHERE jobtype='nurse' );
fiddle
This is how I understood it; what you said:
their salary is less than all nurses
probably means their average salary, so I used it in a subquery.
select e.fname, e.lname, e.salary, e.ssn
from employee e join nurse n on e.ssn = n.ssn
where e.joptype = 'nurse'
and n.shift = 'morning'
and salary < (select avg(salary) --> average salary of all nurses
from employee
where joptype = 'nurse'
);
(Just a remark: is it really joptype? Is it not a job?)
Maybe you could try it like here:
select
Fname, Lname, salary, ssn
from
Employee
where
joptype='nurse' AND
salary <= ( Select MIN(salary)
From Employee E
Inner join Nurses N ON(E.Ssn=N.Ssn)
Where E.joptype = 'nurse' AND
N.shift = 'morning'
)
... this will give you the list of nurses with salary less than or equal to the minimum salary among morning nurses..
In the employees table I have id_departament
And I can`t figure it out how to extract the avg salary for every employee
Use window functions if your RDBMS supports them:
select *
from (
select e.*, avg(salary) over(partition by id_department) avg_salary_dept
from employee e
) t
where salary > avg_salary_dept
Alternatively, you can join the table with an aggregate query that computes the average salary per department:
select e.*
from employee e
inner join (
select id_department, avg(salary) avg_salary_dept
from employee
group by id_department
) a on e.id_department = a.id_department and e.salary > a.avg_salary_dept
I am trying to compile a query which gives me the highest salary per each department and for each unique employee. The complexity is that 1 employee can be part of multiple departments.
In case the same employee has the highest salary in several departments, only the department with a lower salary should show. This is my start but I am not sure how to continue from here:
select max(salary) as salary, dd.dept_name,d.emp_no
from salaries s
inner join dept_emp d on
s.emp_no=d.emp_no
inner join departments dd on
d.dept_no=dd.dept_no
group by 2,3;
My output is:
What should I modify from here?
For an employee, you seem to only want to include the department with the smallest salary. I would recommend using window functions:
select s.*
from (select s.*,
rank() over (partition by dept_name order by salary desc) as seqnum_d
from (select s.*, d.dept_name,
rank() over (partition by dept_name order by salary) as seqnum_ed
from salaries s join
dept_emp d
on s.emp_no = d.emp_no join
departments dd
d.dept_no = dd.dept_no
) s
where seqnum_ed = 1
) s
where seqnum_d = 1;
Something like this?
select m.salary, m.emp_no, salary.dept_name from salary,
(select emp_no, min(salary) salary from salary group by emp_no) m
where
m.emp_no=salary.emp_no and m.salary=salary.salary;
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
O community, do you know how I could select the department_ID, and lowest salary of the department with the highest average salary? Or how to eliminate the'ORA-00934: group function not allowed here' issue? Would I need to use two subqueries?
So far, this is what I've come up with, trying to get the department_ID of the highest paid department:
SELECT department_ID, MIN(salary
FROM employees
WHERE department_ID = (SELECT department_ID
FROM employees WHERE salary = MAX(salary));
Thank you, your assistance is greatly appreciated.
I can't test this, but it should work:
;WITH DepartmentsSalary AS
(
SELECT department_ID, AVG(Salary) AvgSalary, MIN(Salary) MinSalary
FROM employees
GROUP BY department_ID
)
SELECT department_ID, MinSalary
FROM ( SELECT department_ID, AvgSalary, MAX(AvgSalary) OVER() MaxSalary, MinSalary
FROM DepartmentsSalary) D
WHERE MaxSalary = AvgSalary
You can use join (then you have just one sub query)
select e1.department_ID, min(e1.salary)
from employees e1
join (
select avg_query.department_ID, max(avg_query.avg_value)
from (
select department_ID, avg(salary) as avg_value
from employees
group by department_ID
) avg_query
) e2 on e2.department_ID = e1.department_ID
;
First sub-query returned average salary for all departments
Next sub-query based on first sub-query returned highest average
salary and related department_ID
Main query returned min salary for department_ID with highest average
salary