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.