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. Basic Join
  4. Top Competitors
  5. Discussions

Top Competitors

Sort by

recency

|

2522 Discussions

|

  • maciejstawiarz1
     + 0 comments

    ORACLE

    Select
    h.hacker_id,
    h.name
From hackers h
Join submissions s
    on h.hacker_id = s.hacker_id
Join challenges c
    on s.challenge_id = c.challenge_id
Join difficulty d
    on c.difficulty_level = d.difficulty_level
Where s.score = d.score
Group by h.hacker_id, h.name
Having count(distinct(s.challenge_id)) > 1
Order by count(distinct(c.challenge_id)) desc, h.hacker_id asc;

  • allanmaakh7
     + 0 comments

    select s.hacker_id , h.name from submissions s join challenges c on s.challenge_id = c.challenge_id join difficulty d on c.difficulty_level = d.difficulty_level join hackers h on s.hacker_id = h.hacker_id where s.score = d.score group by s.hacker_id, h.name having count() > 1 order by count() desc, s.hacker_id;

  • f20202283h
     + 0 comments

    select s.hacker_id, name from submissions s join hackers h on s.hacker_id = h.hacker_id join Challenges c on s.challenge_id = c.challenge_id join Difficulty d on c.difficulty_level = d.difficulty_level group by s.hacker_id, name having sum(case when s.score = d.score then 1 else 0 end)>1 order by sum(case when s.score = d.score then 1 else 0 end) desc, s.hacker_id

  • Thilak05
     + 0 comments
    SELECT h.hacker_id, h.name
FROM hackers h
JOIN submissions s ON h.hacker_id = s.hacker_id
JOIN challenges c ON s.challenge_id = c.challenge_id
JOIN difficulty d ON c.difficulty_level = d.difficulty_level
WHERE s.score = d.score
GROUP BY h.hacker_id, h.name
HAVING COUNT(DISTINCT c.challenge_id) > 1
ORDER BY COUNT(DISTINCT c.challenge_id) DESC, h.hacker_id ASC;

  • krishna1627
     + 0 comments

    my sql : SELECT aaa.hacker_id, b.name FROM ( SELECT a.hacker_id, COUNT(DISTINCT a.challenge_id) AS totchal FROM submissions a LEFT JOIN challenges b ON a.challenge_id = b.challenge_id LEFT JOIN difficulty c ON b.difficulty_level = c.difficulty_level WHERE a.score = c.score GROUP BY a.hacker_id ) aaa LEFT JOIN hackers b ON aaa.hacker_id = b.hacker_id WHERE totchal > 1 ORDER BY totchal DESC, aaa.hacker_id ASC;