강의/SQL
3주차 SQL 과제
MotherCarGasoline
2022. 2. 20. 23:42
select e.enrolled_id,
e.user_id,
count(*) as cnt
from enrolleds e
inner join enrolleds_detail ed on e.enrolled_id = ed.enrolled_id
where ed.done = 1
group by e.enrolled_id, e.user_id
order by cnt desc
select e.enrolled_id, e.user_id, count(*) as max_count from enrolleds e
inner join enrolleds_detail ed
on e.enrolled_id = ed.enrolled_id
where ed.done = 1
group by e.enrolled_id, e.user_id
order by max_count desc
아직 다행히 기초적인 것들이 더 많아 수월하게 따라 갈 수 있다.