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Here's mine in Oracle, simplest way it seems : Hacker_id in challenges correspond to hackers who created them so watch out !
select h.hacker_id, h.name--,count(distinct s.submission_id) from hackers h join Submissions s on h.hacker_id = s.hacker_id join Difficulty d on s.score = d.score join Challenges c on s.challenge_id = c.challenge_id and d.difficulty_level = c.difficulty_level group by h.hacker_id, h.name having count(s.submission_id) > 1 order by count(s.submission_id) desc, hacker_id asc;
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You are viewing a single comment's thread. Return to all comments →
Here's mine in Oracle, simplest way it seems : Hacker_id in challenges correspond to hackers who created them so watch out !