I have to write query that returns all employees from departments with average salary smaller than 12000.
I wrote down this query
SELECT DEPARTMENT_ID, FIRST_NAME, LAST_NAME
FROM EMPLOYEES
GROUP BY DEPARTMENT_ID,FIRST_NAME,LAST_NAME
HAVING AVG(SALARY) < 12000
ORDER BY LAST_NAME ASC;
but it doesn't seem to work. Instead of taking all employees from those departments it takes only ones with salary smaller than 12000.
How to write this query properly?
Thanks in advance.
You need the average salary for the department. I would use a window function:
SELECT DEPARTMENT_ID, FIRST_NAME, LAST_NAME
FROM (SELECT e.*, AVG(SALARY) OVER (PARTITION BY DEPARTMENT_ID) as avg_salary_dept
FROM EMPLOYEES e
) e
WHERE avg_salary_dept < 12000
ORDER BY LAST_NAME ASC;
You need subquery :
SELECT e.DEPARTMENT_ID, e.FIRST_NAME, e.LAST_NAME
FROM EMPLOYEES e
WHERE e.DEPARTMENT_ID IN (SELECT e1.DEPARTMENT_ID
FROM EMPLOYEES e1
GROUP BY e1.DEPARTMENT_ID
HAVING AVG(e1.SALARY) < 12000
)
ORDER BY e.LAST_NAME ASC;
You can use correlated subquery
SELECT e.DEPARTMENT_ID, e.FIRST_NAME, e.LAST_NAME
FROM EMPLOYEES a
WHERE exists (SELECT 1 FROM EMPLOYEES b
where a.department_id=b.department_id
HAVING AVG(b.SALARY) < 12000
);
Try it..
SELECT e.DEPARTMENT_ID, e.FIRST_NAME, e.LAST_NAME
from employee e where e.DEPARTMENT_ID in
(select e.DEPARTMENT_ID from employee e
group by e.DEPARTMENT_ID
having avg(salary)<3000)
Related
I am using an Oracle Developer Database and I have the two tables, Employees and Departments :
I am supposed to find (using a correlated subquery) for each department all the employees, which have a salary within the range of the salary-standard deviation of every single department.
This is what I have tried so far:
SELECT d.department_name, e.employee_id, e.last_name, AVG(e.salary), e.salary
FROM hr.employees e
JOIN hr.departments d
on e.department_id = d.department_id
WHERE e.salary IN
(SELECT salary
FROM hr.employees
WHERE salary >
(SELECT ROUND(AVG(e.salary)-STDDEV(e.salary),2)
FROM hr.employees) AND salary < (SELECT ROUND(AVG(e.salary) + STDDEV(e.salary),2) FROM hr.employees)
GROUP BY d.department_name,
e.employee_id, e.last_name, e.salary;
Even though this query has the correct syntax, it does not show any result.
This other (partial) approach works but how can I query at the same time the employee table?
SELECT d.department_name, e.department_id, ROUND(AVG(e.salary) + STDDEV(e.salary), 2)
AS standard_deviation_max,
ROUND(AVG(e.salary) - STDDEV(e.salary), 2)
AS standard_deviation_min
FROM hr.employees e
JOIN hr.departments d
on e.department_id = d.department_id
GROUP BY department_name, e.department_id;
Since I am new to SQL, I would really appreciate any hint. Thank you in advance
You can use your query to find the employee details and then correlate the salary to the average ± the standard deviation:
SELECT d.department_name,
e.employee_id,
e.last_name,
e.salary
FROM hr.employees e
JOIN hr.departments d
on e.department_id = d.department_id
WHERE EXISTS (
SELECT 1
FROM hr.employees x
HAVING e.salary BETWEEN AVG(e.salary) + STDDEV(e.salary)
AND AVG(e.salary) + STDDEV(e.salary)
);
or, you can use analytic functions:
SELECT department_name,
employee_id,
last_name,
salary
FROM (
SELECT d.department_name,
e.employee_id,
e.last_name,
e.salary,
AVG(e.salary) OVER () AS avg_salary,
STDDEV(e.salary) OVER () AS stddev_salary
FROM hr.employees e
JOIN hr.departments d
on e.department_id = d.department_id
)
WHERE salary BETWEEN avg_salary - stddev_salary
AND avg_salary + stddev_salary;
I am stuck on a school project. I need the second-highest salary per department, but without having the third, fourth etc. I am using Oracle SQL. and I need to do it with subquery.
This is the only code I can come up with so far. I can filter all salaries after the highest, but I cant filter only the second highest. I have two statements, that produce the same result at the end.
select e.department_id, d.department_name, e.first_name, e.last_name, e.salary
from employees e inner join departments d
on (e.department_id = d.department_id)
where e.salary < (select max(salary)
from employees e
where e.department_id = d.department_id
order by d.department_id
offset 1 row fetch next 2 row only)
order by e.department_id;
Or
select department_id, first_name, last_name, salary
from employees
where salary < any (select max(salary)
from employees
group by department_id)
order by department_id;
I have tried this but obviously it will only give mt the average salary for all the departments not each job classification
SELECT e.department_id, e.job_id, (SELECT avg(salary)
FROM employees)
FROM employees e
GROUP BY department_id, job_id;
Do you want the department's avg salary? Do a correlated sub-query:
SELECT e.department_id, e.job_id, (SELECT avg(salary)
FROM employees e2
where e2.department_id = e.department_id)
FROM employees e
I suppose you want this (if your aim is get average salary for couple department_id / job_id):
SELECT e.department_id, e.job_id, avg(e.salary)
FROM employees e
GROUP BY e.department_id, e.job_id;
If you want average only for department_id you can write:
SELECT e.department_id, e.job_id,
(SELECT avg(e1.salary)
FROM employees e1
where e1.department_id = e.department_id)
FROM employees e
GROUP BY e.department_id, e.job_id;
I found a couple of SQL tasks on Hacker News today, however I am stuck on solving the second task in Postgres, which I'll describe here:
You have the following, simple table structure:
List the employees who have the biggest salary in their respective departments.
I set up an SQL Fiddle here for you to play with. It should return Terry Robinson, Laura White. Along with their names it should have their salary and department name.
Furthermore, I'd be curious to know of a query which would return Terry Robinsons (maximum salary from the Sales department) and Laura White (maximum salary in the Marketing department) and an empty row for the IT department, with null as the employee; explicitly stating that there are no employees (thus nobody with the highest salary) in that department.
Return one employee with the highest salary per dept.
Use DISTINCT ON for a much simpler and faster query that does all you are asking for:
SELECT DISTINCT ON (d.id)
d.id AS department_id, d.name AS department
,e.id AS employee_id, e.name AS employee, e.salary
FROM departments d
LEFT JOIN employees e ON e.department_id = d.id
ORDER BY d.id, e.salary DESC;
->SQLfiddle (for Postgres).
Also note the LEFT [OUTER] JOIN that keeps departments with no employees in the result.
This picks only one employee per department. If there are multiple sharing the highest salary, you can add more ORDER BY items to pick one in particular. Else, an arbitrary one is picked from peers.
If there are no employees, the department is still listed, with NULL values for employee columns.
You can simply add any columns you need in the SELECT list.
Find a detailed explanation, links and a benchmark for the technique in this related answer:
Select first row in each GROUP BY group?
Aside: It is an anti-pattern to use non-descriptive column names like name or id. Should be employee_id, employee etc.
Return all employees with the highest salary per dept.
Use the window function rank() (like #Scotch already posted, just simpler and faster):
SELECT d.name AS department, e.employee, e.salary
FROM departments d
LEFT JOIN (
SELECT name AS employee, salary, department_id
,rank() OVER (PARTITION BY department_id ORDER BY salary DESC) AS rnk
FROM employees e
) e ON e.department_id = d.department_id AND e.rnk = 1;
Same result as with the above query with your example (which has no ties), just a bit slower.
This is with reference to your fiddle:
SELECT * -- or whatever is your columns list.
FROM employees e JOIN departments d ON e.Department_ID = d.id
WHERE (e.Department_ID, e.Salary) IN (SELECT Department_ID, MAX(Salary)
FROM employees
GROUP BY Department_ID)
EDIT :
As mentioned in a comment below, if you want to see the IT department also, with all NULL for the employee records, you can use the RIGHT JOIN and put the filter condition in the joining clause itself as follows:
SELECT e.name, e.salary, d.name -- or whatever is your columns list.
FROM employees e RIGHT JOIN departments d ON e.Department_ID = d.id
AND (e.Department_ID, e.Salary) IN (SELECT Department_ID, MAX(Salary)
FROM employees
GROUP BY Department_ID)
This is basically what you want. Rank() Over
SELECT ename ,
departments.name
FROM ( SELECT ename ,
dname
FROM ( SELECT employees.name as ename ,
departments.name as dname ,
rank() over (
PARTITION BY employees.department_id
ORDER BY employees.salary DESC
)
FROM Employees
JOIN Departments on employees.department_id = departments.id
) t
WHERE rank = 1
) s
RIGHT JOIN departments on s.dname = departments.name
Good old classic sql:
select e1.name, e1.salary, e1.department_id
from employees e1
where e1.salary=
(select maxsalary=max(e.salary) --, e. department_id
from employees e
where e.department_id = e1.department_id
group by e.department_id
)
Table1 is emp - empno, ename, sal, deptno
Table2 is dept - deptno, dname.
Query could be (includes ties & runs on 11.2g):
select e1.empno, e1.ename, e1.sal, e1.deptno as department
from emp e1
where e1.sal in
(SELECT max(sal) from emp e, dept d where e.deptno = d.deptno group by d.dname)
order by e1.deptno asc;
SELECT
e.first_name, d.department_name, e.salary
FROM
employees e
JOIN
departments d
ON
(e.department_id = d.department_id)
WHERE
e.first_name
IN
(SELECT TOP 2
first_name
FROM
employees
WHERE
department_id = d.department_id);
`select d.Name, e.Name, e.Salary from Employees e, Departments d,
(select DepartmentId as DeptId, max(Salary) as Salary
from Employees e
group by DepartmentId) m
where m.Salary = e.Salary
and m.DeptId = e.DepartmentId
and e.DepartmentId = d.DepartmentId`
The max salary of each department is computed in inner query using GROUP BY. And then select employees who satisfy those constraints.
Assuming Postgres
Return highest salary with employee details, assuming table name emp having employees department with dept_id
select e1.* from emp e1 inner join (select max(sal) avg_sal,dept_id from emp group by dept_id) as e2 on e1.dept_id=e2.dept_id and e1.sal=e2.avg_sal
Returns one or more people for each department with the highest salary:
SELECT result.Name Department, Employee2.Name Employee, result.salary Salary
FROM ( SELECT dept.name, dept.department_id, max(Employee1.salary) salary
FROM Departments dept
JOIN Employees Employee1 ON Employee1.department_id = dept.department_id
GROUP BY dept.name, dept.department_id ) result
JOIN Employees Employee2 ON Employee2.department_id = result.department_id
WHERE Employee2.salary = result.salary
SQL query:
select d.name,e.name,e.salary
from employees e, depts d
where e.dept_id = d.id
and (d.id,e.salary) in
(select dept_id,max(salary) from employees group by dept_id);
Take look at this solution
SELECT
MAX(E.SALARY),
E.NAME,
D.NAME as Department
FROM employees E
INNER JOIN DEPARTMENTS D ON D.ID = E.DEPARTMENT_ID
GROUP BY D.NAME
I am writing a query to find employees who earn greater than the average salary within their department. I need to display the employee ID, salary, department id, and average salary of that department.
I have a query that just almost works but it keeps giving me "ORA-00904: "AVG_SAL": invalid identifier" errors. Am I doing this correctly. Why am i getting this invalid identifier error?
SELECT employee_id, salary, department_id,
(SELECT ROUND(AVG(salary),2)
FROM employees e_inner
WHERE e_inner.department_id = e.department_id) AS avg_sal
FROM employees e
WHERE salary > avg_sal
ORDER BY avg_sal DESC
More efficient to use analytics:
select employee_id, salary, department_id, avg_sal
from
(
SELECT employee_id, salary, department_id,
round(avg(salary) over (partition by department_id), 2) avg_sal
from emp
)
where salary > avg_sal
order by avg_sal desc
I don't believe you can refer to a column alias (avg_sal in this case) in a WHERE clause.
You'll need to repeat that inner query, i.e.:
SELECT employee_id, salary, department_id,
(SELECT ROUND(AVG(salary),2)
FROM employees e_inner
WHERE e_inner.department_id = e.department_id) AS avg_sal
FROM employees e
WHERE salary >
(SELECT ROUND(AVG(salary),2)
FROM employees e_inner
WHERE e_inner.department_id = e.department_id)
ORDER BY avg_sal DESC
Not great, with those two inner queries, but that's the most-straightforward way to correct the error.
Update: Haven't tested this, but try the following:
SELECT e.employee_id, e.salary, e.department_id, b.avg_sal
FROM employees e
INNER JOIN
(SELECT department_id, ROUND(AVG(salary),2) AS avg_sal
FROM employees
GROUP BY department_id) e_avg ON e.department_id = e_avg.department_id AND e.salary > e_avg.avg_sal
ORDER BY e_avg.avg_sal DESC
You could rewrite it as a join:
SELECT e1.employee_id
, e1.salary
, e1.department_id
, ROUND(AVG(e2.salary),2) as Avg_Sal
FROM employees e
JOIN employees e2
ON e2.department_id = e.department_id
GROUP BY
e1.employee_id
, e1.salary
, e1.department_id
HAVING e1.salary > ROUND(AVG(e2.salary),2)
Or a subquery:
SELECT *
FROM (
SELECT employee_id
, salary
, department_id
, (
SELECT ROUND(AVG(salary),2)
FROM employees e_inner
WHERE e_inner.department_id = e.department_id
) AS avg_sal
FROM employees e
) as SubqueryAlias
WHERE salary > avg_sal
select *
from employees e
join(
select Round(avg(salary)) AvgSal,department_id,department_name as dept_name
from employees join departments
using (department_id)
group by department_id,department_name
) dd
using(department_id)
where e.salary > dd.AvgSal;
another solution
select *
from employees e,
(
select
department_id,
avg(salary) avg_sal
from employees
group by department_id
) e1
where e.department_id=e1.department_id
and e.salary > e1.avg_sal