• + 0 comments

    select c.company_code,c.founder, count(distinct l.lead_manager_code), count(distinct s.senior_manager_code), count(distinct m.manager_code), count(distinct e.employee_code) from company as c join lead_manager as l using (company_code) join senior_manager as s using (lead_manager_code) join manager as m using (senior_manager_code) join employee as e using (manager_code) GROUP BY C.company_code , C.FOUNDER ORDER BY C.company_code;