[프로그래머스] 즐겨찾기가 가장 많은 식당 정보 출력 - GROUP BY , JOIN

https://school.programmers.co.kr/learn/courses/30/lessons/131123

프로그래머스

 

 

-- 코드를 입력하세요
SELECT a.FOOD_TYPE, a.REST_ID, a.REST_NAME, a.FAVORITES
FROM REST_INFO a 
JOIN (SELECT FOOD_TYPE, MAX(FAVORITES) AS FAVORITES
      FROM REST_INFO
      GROUP BY FOOD_TYPE
        ) b 
        ON a.FOOD_TYPE = b.FOOD_TYPE AND a.FAVORITES = b.FAVORITES
ORDER BY FOOD_TYPE DESC