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  1. Prepare
  2. SQL
  3. Advanced Select
  4. New Companies
  5. Discussions

New Companies

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3284 Discussions

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  • simonjaramillor
     + 0 comments

    SELECT comp.company_code, comp.founder, (SELECT COUNT() FROM lead_manager WHERE company_code = comp.company_code) AS total_leads, (SELECT COUNT() FROM senior_manager WHERE company_code = comp.company_code) AS total_seniors, (SELECT COUNT() FROM manager WHERE company_code = comp.company_code) AS total_managers, (SELECT COUNT() FROM employee WHERE company_code = comp.company_code) AS total_employees FROM company AS comp ORDER BY comp.company_code ASC;

  • pfms_vikask
     + 0 comments

    For MS SQL SErver

    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 c JOIN lead_manager l on l.company_code = c.company_code JOIN Senior_Manager s on l.lead_manager_code = s.lead_manager_code join Manager m on s.Senior_Manager_code = m.Senior_Manager_code JOIN Employee e on m.Manager_code = e.Manager_code

    group by c.company_code, c.founder order by c.company_code

  • stochasticboy
     + 0 comments

    select company.company_code, company.founder, count(distinct(employee.lead_manager_code)), count(distinct(employee.senior_manager_code)), count(distinct(employee.manager_code)), count(distinct(employee.employee_code)) from company inner join employee on company.company_code=employee.company_code

    group by company.company_code, company.founder order by company.company_code;

  • rjd19972
     + 0 comments

    SELECT c.company_code, c.founder, Count(Distinct(lm.lead_manager_code)), Count(Distinct(sm.senior_manager_code)), Count(Distinct(m.manager_code)), Count(Distinct(e.employee_code)) from Company c join Lead_Manager lm on c.company_code = lm.company_code join Senior_Manager sm ON c.company_code = sm.company_code join Manager m ON c.company_code = m.company_code join Employee e ON c.company_code = e.company_code GROUP BY c.company_code, c.founder ORDER BY c.company_code;

  • shreyas_auti_201
     + 0 comments

    select a.company_code, a.founder, count(distinct(b.lead_manager_code)), count(distinct(b.senior_manager_code)), count(distinct(b.manager_code)), count(distinct(b.employee_code)) from company a join employee b on a.company_code=b.company_code group by 1,2 order by 1