data_yep

[문제]  [MySQL]SELECT F1.X , F1.YFROM Functions AS F1 INNER JOIN Functions AS F2 ON F1.X = F2.Y AND F1.Y = F2.XGROUP BY F1.X, F1.YHAVING F1.X = 2ORDER BY F1.X ASC

[문제]  [MySQL]SELECT S.NameFROM Students AS S INNER JOIN Friends AS F ON S.ID = F.ID INNER JOIN Packages AS P1 ON P1.ID = F.ID INNER JOIN Packages AS P2 ON P2.ID = F.Friend_IDWHERE P1.Salary

[문제]  [MySQL]WITH step1 AS ( SELECT ROW_NUMBER() OVER (ORDER BY Start_Date) AS id , LAG(End_Date, 1) OVER (ORDER BY Start_Date) AS last_end , Start_Date , End_Date , LEAD(Start_Date, 1) OVER (ORDER BY Start_Date) AS next_start FROM Projects), step2 AS ( SELECT id , CASE WHEN last_end != Start_Date THEN id WHEN last_end IS NULL THEN ..

[문제]  [MySQL]SELECT hacker_id, name, SUM(score) AS total_scoreFROM (SELECT h.hacker_id, h.name, s.challenge_id, MAX(s.score) AS score FROM Hackers AS h LEFT JOIN Submissions AS s ON h.hacker_id = s.hacker_id GROUP BY h.hacker_id, h.name, s.challenge_id) AS high_scoreGROUP BY hacker_id, nameHAVING SUM(score) != 0ORDER BY total_score DESC, hacker_id ASC

[문제]  [MySQL]WITH challenge_cnts AS ( SELECT h.hacker_id, h.name, COUNT(c.challenge_id) AS cnt FROM Hackers AS h LEFT JOIN Challenges AS c ON h.hacker_id = c.hacker_id GROUP BY h.hacker_id, h.name), same_cnts AS ( SELECT *, COUNT(*) OVER (PARTITION BY cnt) AS same_cnt FROM challenge_cnts)SELECT hacker_id, name, cntFROM same_cntsWHERE same_cnt = 1 OR cnt = (SELECT MAX(cnt)..

[문제]  [MySQL]SELECT w.id, wp.age, w.coins_needed, w.powerFROM Wands AS w LEFT JOIN Wands_Property AS wp ON w.code = wp.codeWHERE wp.is_evil = 0 AND w.coins_needed = (SELECT MIN(w1.coins_needed) FROM Wands AS w1 LEFT JOIN Wands_Property AS w2 ON w1.code = w2.code WHERE w1.power = w.power ..

[문제]  [MySQL]SELECT h.hacker_id , h.nameFROM Submissions AS s LEFT JOIN Hackers AS h ON h.hacker_id = s.hacker_id LEFT JOIN Challenges AS c ON s.challenge_id = c.challenge_id LEFT JOIN Difficulty AS d ON c.difficulty_level = d.difficulty_levelWHERE d.score = s.scoreGROUP BY h.hacker_id, h.nameHAVING COUNT(h.hacker_id) > 1ORDER BY COUNT(h.hacker_id) DESC, h.hacker_id ASC

[문제]  [MySQL]SELECT CASE WHEN G.Grade >= 8 THEN S.Name ELSE Null END AS Name , G.Grade , S.MarksFROM Students AS S INNER JOIN Grades AS G ON S.Marks BETWEEN G.Min_Mark AND G.Max_MarkORDER BY G.Grade DESC, Name ASC, S.Marks ASC

[문제]  [MySQL]SELECT CO.Continent , FLOOR(AVG(C.Population))FROM CITY AS C INNER JOIN COUNTRY AS CO ON C.CountryCode = CO.CodeGROUP BY CO.Continent

[문제]  [MySQL]SELECT C.NameFROM CITY AS C INNER JOIN COUNTRY AS CO ON C.CountryCode = CO.CODEWHERE CO.CONTINENT = 'Africa'