как выбрать максимальную (зарплату) сотрудника в каждом отделе с employee_id и emp_name

Question

Я хочу выбрать emp_id, департамент_ид, макс. (Зарплата) каждого отдела, но я использую группу по отделу_иде, и в ней есть ошибка ora-00979

3 столбец находится в той же таблице (сотрудники)

Как это исправить

select  department_id, employee_id as "ID",first_name || ' ' || last_name as "Name",max(salary)as "SALARY"
from EMPLOYEES
group by  department_id
order by department_id;

Answer 1

используйте запрос ниже:

SELECT e.department_id,e.employee_id,
  e.first_name||' '|| e.last_name AS emp_name,t1.max_sal
FROM
  (SELECT department_id,
    MAX(salary) AS max_sal
  FROM employees
  GROUP BY department_id
  ) t1
JOIN employees e
ON t1.department_id = e.department_id
AND t1.max_sal = e.salary
order by e.department_id;

Answer 2

Попробуйте это:

SELECT department_id, salary AS "Salary", employee_id AS "ID", first_name || ' ' || last_name AS "Name" FROM employees 
WHERE salary = (SELECT MAX(salary) FROM employees) GROUP BY department_id;

Я надеюсь, что это работает.:)

Answer 3

Вы можете использовать keep:

select  department_id,
        max(employee_id) keep (dense_rank first order by salary desc) as "ID",
        max(first_name || ' ' || last_name) keep (dense_rank first order by salary desc, employee_id desc) as "Name", 
        max(salary) as "SALARY"
from employees e
group by  department_id
order by department_id;

Answer 4

Ошибка в том, что employee_id отсутствует в группе.

Возможное решение:

select department_id, ID, Name, SALARY
from (
    select distinct department_id, 
        first_value(employee_id) over (partition by department_id order by salary desc) as ID,
        first_value(first_name || ' ' || last_name) over (partition by department_id order by salary desc) as Name,
        first_value(salary) over (partition by department_id order by salary desc)as SALARY
    from EMPLOYEES
)
order by department_id;

Answer 5

Используйте подзапрос с отделом_id и максимальной зарплатой, а затем присоединитесь к основной таблице:

select 
  e.department_id, 
  t.employee_id as id,
  t.first_name || ' ' || t.last_name as name,
  e.maxsalary 
from (
  select 
    department_id, 
    max(salary) as maxsalary 
  from 
    EMPLOYEES
  group by 
    department_id 
) e
inner join 
  EMPLOYEES t 
on 
  t.department_id = e.department_id and t.salary = e.maxsalary
order by e.department_id;

См. демо

СОТРУДНИКИ

EMPLOYEE_ID DEPARTMENT_ID   SALARY  FIRST_NAME  LAST_NAME
    1          1            10000       A           B
    2          1            20000       C           D
    3          1            150000      E           F
    4          2            12000       G           H
    5          2            10000       I           J
    6          3            20000       K           L
    7          4            11000       M           N
    8          4            11000       O           P
    9          4            11000       Q           R
    10         4            10000       S           T

Результат

DEPARTMENT_ID   ID  NAME    MAXSALARY
    1            3  E F      150000
    2            4  G H       12000
    3            6  K L       20000
    4            7  M N       11000
    4            8  O P       11000
    4            9  Q R       11000