We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. SQL
  3. Advanced Select
  4. New Companies
  5. Discussions

New Companies

Sort 1475 Discussions, By:

  • Karthiksk
     + 107 comments

    This the solution I came up with for MySQL

    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, Lead_Manager l, Senior_Manager s, Manager m, Employee e 
where c.company_code = l.company_code 
    and l.lead_manager_code=s.lead_manager_code 
    and s.senior_manager_code=m.senior_manager_code 
    and m.manager_code=e.manager_code 
group by c.company_code order by c.company_code;

  • phyolintun87
     + 20 comments

    Fixed version

    select c.company_code, 
    c.founder, 
    count(distinct e.lead_manager_code), 
    count(distinct e.senior_manager_code), 
    count(distinct e.manager_code), 
    count(distinct e.employee_code)
from company c
    inner join employee e on e.company_code = c.company_code
group by c.company_code,c.founder
order by c.company_code;

  • Rohan_Warang
     + 22 comments

    Getting confused about using Joins or not is quite obvious.Below is my code which does not use any joins,but still gets the job done!!

    select company_code, founder,
(select count(distinct lead_manager_code) from Lead_Manager where company_code=c.company_code),
(select count(distinct senior_manager_code) from Senior_Manager where company_code=c.company_code),
(select count(distinct manager_code) from Manager where company_code=c.company_code),
(select count(distinct employee_code) from Employee where company_code=c.company_code)
from Company c
order by company_code;

  • sakshidgoel
     + 4 comments

    Solution 1

    This is the solution I came up with for MySQL: 

    SELECT c.company_code, c.founder, COUNT(DISTINCT e.lead_manager_code), COUNT(DISTINCT e.senior_manager_code), COUNT(DISTINCT e.manager_code), COUNT(DISTINCT e.employee_code)
FROM Company AS c
INNER JOIN Employee AS e
ON c.company_code = e.company_code
GROUP BY c.company_code, c.founder
ORDER BY c.company_code;

    This code only uses the two tables 'Company' and 'Employee', instead of using all the 5 tables provided.

  • effi818
     + 4 comments

    The sorting requirement is poorly stated. The descriptions says 'ascending', however the output example isn't sorted by an ascending order: C1 C2 ...

Load more conversations