데이터 분석 기술 블로그

Solution 1SELECT DISTINCT(CITY)FROM STATIONWHERE CITY NOT REGEXP '^[aeiouAEIOU]' AND CITY NOT REGEXP '[aeiouAEIOU]$'  Solution 2SELECT DISTINCT(CITY)FROM STATIONWHERE CITY REGEXP '^[^aeiouAEIOU]' AND CITY REGEXP '[^aeiouAEIOU]$'

SolutionSELECT query_name, ROUND(SUM(rating / position) / COUNT(query_name), 2) AS quality, ROUND(SUM(CASE WHEN rating  1. rating이 3 이하인 개수를 구해야 하는데 CASE WHEN을 사용할 수 있는지 몰랐었다. 그리고 꼭 END를 뒤에 붙여야 한다.2. WHERE 절 없이 제출했더니 query_name이 null인 경우는 계산이 안되게 해야 했다.. 근데 문제에 그런 말이 없는데.. Test Case에도 없고.. 디버깅한 것으로 만족해야겠다.

Solution1SELECT DISTINCT(CITY)FROM STATIONWHERE CITY REGEXP '^[^aeiouAEIOU]' OR CITY REGEXP '[^aeiouAEIOU]$'; Solution 2SELECT DISTINCT(CITY)FROM STATIONWHERE CITY NOT REGEXP '^[aeiouAEIOU]' OR CITY NOT REGEXP '[aeiouAEIOU]$';

SolutionSELECT contest_id, ROUND(COUNT(contest_id) * 100 / (SELECT COUNT(*) FROM Users), 2) AS percentageFROM RegisterGROUP BY contest_idORDER BY percentage DESC, contest_id ASC; Users에서 총개수를 구해야 한다는 것은 알고 있었는데 JOIN이 아니라 어떤 식으로 구할지를 몰라서 확인해 보니 총 개수 구하는 부분에서 SQL문을 넣어서 사용하면 되었다.

SolutionSELECT DISTINCT(CITY)FROM STATIONWHERE CITY REGEXP '[^aeiouAEIOU]$';

SolutionSELECT p.project_id, ROUND(AVG(e.experience_years), 2) AS average_yearsFROM Project pLEFT JOIN Employee eON p.employee_id = e.employee_idGROUP BY p.project_id;

SolutionSELECT DISTINCT(CITY)FROM STATIONWHERE CITY REGEXP '^[^aeiouAEIOU]';

SolutionSELECT p.product_id, ROUND(CASE WHEN SUM(u.units) IS NULL THEN 0 ELSE SUM(p.price * u.units) / SUM(u.units) END, 2) AS average_priceFROM Prices pLEFT JOIN UnitsSold uOn p.product_id = u.product_id AND u.purchase_date BETWEEN p.start_date AND p.end_dateGROUP BY p.product_id; Solution 2SELECT p.product_id, ROUND(COALESCE(SUM(p.price * u.units) / SUM(u.uni..