Oracle Show All Employees with Greater Than Average Salary of Their Department

Oracle show all employees with greater than average salary of their department

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

Write a SQL code to select all employees with salary greater than the average department salary and also greater than $30K

This may work

select s1.employeeid, s1.departmentid, s1.salary
from salary as s1
  inner join
  (select departmentid, avg(salary) as avgsalary
    from salary 
    group by departmentid) as s2   
  on s1.departmentid = s2.departmentid
where salary > avgsalary and salary > 30000
;

SQL - Trying to get employees with greater than average department salary

Window functions are probably the best solution -- as in @zealous's answer. However, a subquery is also very reasonable. The idea is:

select e.first_name, e.salary, e.department_id
from employees as e
where e.salary > (select avg(e2.salary)
                  from employees e2
                  where e2.department_id = e.department_id
                 );

Here is a db<>fiddle.

Key ideas:

• You don't need to join to departments. Just because you have the idea doesn't mean the join is necessary.
• The GROUP BY is unnecessary in the subquery. The correlation clause takes care of that.

How to check if a value is larger than the average of these values?

Use AVG as an analytic function:

SELECT full_name
FROM   (
  SELECT e.first_name || ' ' ||e.last_name AS full_name,
         e.salary,
         AVG( e.salary ) OVER ( PARTITION BY l.country_id ) AS avg_salary
  FROM   hr.employees e
         INNER JOIN HR.departments d
         ON e.department_id = d.department_id
         INNER JOIN HR.locations l
         ON d.location_id = l.location_id
)
WHERE  salary > avg_salary;

How to select employees with salary less then the average of their departament using SQL analytical functions?

Use a window function and a subquery:

select e.*
from (select e.*, 
             avg(salary) over (partition by id_dep) as avg_salary
      from employees e
     ) e
where salary < avg_salary;

You cannot use window functions in where or having clauses. Some databases support a qualify clause that filters on window functions; but Oracle does not.

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