Can I group by in SQL query with window function?

I need to get employees with smallest salary in their departments I did it using anti join.

     select emp.employee_id,emp.last_name,emp.salary,emp.department_id
     from employees emp
     left join employees sml 
     on sml.department_id = emp.department_id and sml.salary < emp.salary
     where sml.employee_id is null and emp.department_id is not null

But I’ve been told that it’s possible to do it using window function using one select. However I can’t group it by department_id and use it at the same time. Is that a bug or me being stupid?

     SELECT  department_id,
     min(salary) OVER (partition by department_id)  as minsalary
     FROM employees;
     GROUP BY department_id

SQL Developer says 00979. 00000 – “not a GROUP BY expression”

Answer

If you run your second query without the group by – which you may have already tried, from the extra semicolon in what you posted – you’ll see that you get one row for every employee, each showing the minimum salary in their department. That minimum is the analytic min() because it has a window clause. The PARTITION BY is the equivalent of a GROUP BY, but without the aggregation over the whole result set.

The simplest way to get the same result (almost) is to use the RANK() analytic function instead, which ranks the values based on the partition and order you supply, while allowing for ties:

SELECT employee_id, last_name, salary, department_id,
  RANK() OVER (PARTITION BY department_id ORDER BY salary) AS rnk
FROM employees
ORDER BY department_id, rnk;

EMPLOYEE_ID LAST_NAME                     SALARY DEPARTMENT_ID        RNK
----------- ------------------------- ---------- ------------- ----------
        200 Whalen                          4400            10          1
        202 Fay                             6000            20          1
        201 Hartstein                      13000            20          2
        119 Colmenares                      2500            30          1
        118 Himuro                          2600            30          2
        117 Tobias                          2800            30          3
        116 Baida                           2900            30          4
        115 Khoo                            3100            30          5
        114 Raphaely                       11000            30          6
...
        102 De Haan                        17000            90          1
        101 Kochhar                        17000            90          1
        100 King                           24000            90          3
...

For departments 20 and 30 you can see the row ranked 1 is the lowest salary. For department 90 there are two employees ranked 1, because they have the same lowest salary.

You can use that as an inline view and select just those rows ranked number 1:

SELECT employee_id, last_name, salary, department_id
FROM (
  SELECT employee_id, last_name, salary, department_id,
    RANK() OVER (PARTITION BY department_id ORDER BY salary) AS rnk
  FROM employees
)
WHERE rnk = 1
ORDER BY department_id;

EMPLOYEE_ID LAST_NAME                     SALARY DEPARTMENT_ID
----------- ------------------------- ---------- -------------
        200 Whalen                          4400            10
        202 Fay                             6000            20
        119 Colmenares                      2500            30
        203 Mavris                          6500            40
        132 Olson                           2100            50
        107 Lorentz                         4200            60
        204 Baer                           10000            70
        173 Kumar                           6100            80
        101 Kochhar                        17000            90
        102 De Haan                        17000            90
        113 Popp                            6900           100
        206 Gietz                           8300           110
        178 Grant                           7000              

13 rows selected. 

If you didn’t have to worry about ties there is an even simpler alternative, but it ins’t appropriate here.

Notice that this gives you one more row than your original query. You are joining on sml.department_id = emp.department_id. If the department ID is null, as it is for employee 178, that join fails because you can’t compare null to null with equality tests. Because this solution doesn’t have a join, that doesn’t apply, and you see that employee in the results.