You are viewing a single comment's thread. Return to all comments →
select s.hacker_id , h.name from submissions s inner join hackers h on s.hacker_id = h.hacker_id inner join challenges c on s.challenge_id = c.challenge_id inner join difficulty d on c.difficulty_level = d.difficulty_level where s.score = d.score group by s.hacker_id,h.name having count(h.name) > 1 order by count(h.name) desc , s.hacker_id asc
Seems like cookies are disabled on this browser, please enable them to open this website
Top Competitors
You are viewing a single comment's thread. Return to all comments →