"HR-SQL Top Competitors"의 두 판 사이의 차이

2018년 8월 17일 (금) 01:49 판

1 개요

HR-SQL Top Competitors

https://www.hackerrank.com/challenges/full-score/problem

해커랭크 SQL
문제	DB2	MS SQL	MySQL	Oracle
HR-SQL Basic Join e
HR-SQL Asian Population
HR-SQL African Cities
HR-SQL Average Population of Each Continent
HR-SQL The Report
HR-SQL Top Competitors
HR-SQL Ollivander's Inventory
HR-SQL Challenges
HR-SQL Contest Leaderboard

2 MySQL

SELECT H.hacker_id, H.name
FROM Submissions S
INNER JOIN Challenges C ON S.challenge_id=C.challenge_id
INNER JOIN Difficulty D ON C.difficulty_level=D.difficulty_level 
INNER JOIN Hackers H ON S.hacker_id=H.hacker_id
WHERE S.score=D.score AND C.difficulty_level=D.difficulty_level
GROUP BY H.hacker_id, H.name
HAVING COUNT(S.hacker_id) > 1
ORDER BY COUNT(S.hacker_id) DESC, S.hacker_id

@@ 1번째 줄: / 1번째 줄: @@
 ==개요==
 ;HR-SQL Top Competitors
-*
+* https://www.hackerrank.com/challenges/full-score/problem
 {{HR-SQL 헤더}}
@@ 7번째 줄: / 7번째 줄: @@
 |}
-<source lang='sql'>
+==MySQL==
+<source lang='mysql'>
+SELECT H.hacker_id, H.name
+FROM Submissions S
+INNER JOIN Challenges C ON S.challenge_id=C.challenge_id
+INNER JOIN Difficulty D ON C.difficulty_level=D.difficulty_level
+INNER JOIN Hackers H ON S.hacker_id=H.hacker_id
+WHERE S.score=D.score AND C.difficulty_level=D.difficulty_level
+GROUP BY H.hacker_id, H.name
+HAVING COUNT(S.hacker_id) > 1
+ORDER BY COUNT(S.hacker_id) DESC, S.hacker_id
 </source>