즐겨찾기가 가장 많은 식당 정보 출력하기

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

• 문제 풀이

SELECT F.FOOD_TYPE, I.REST_ID, I.REST_NAME, F.FAVORITES
FROM (SELECT FOOD_TYPE, MAX(FAVORITES) AS "FAVORITES" 
      FROM REST_INFO 
      GROUP BY FOOD_TYPE) F
JOIN REST_INFO I 
ON F.FOOD_TYPE = I.FOOD_TYPE
WHERE F.FAVORITES = I.FAVORITES
ORDER BY F.FOOD_TYPE DESC;

SELECT I.FOOD_TYPE, 
       I.REST_ID, 
       I.REST_NAME, 
       I.FAVORITES
FROM 
    (SELECT DISTINCT FOOD_TYPE, MAX(FAVORITES) AS "FAVORITES"
     FROM REST_INFO
     GROUP BY FOOD_TYPE) F, REST_INFO I
WHERE F.FOOD_TYPE = I.FOOD_TYPE
  AND F.FAVORITES = I.FAVORITES
ORDER BY FOOD_TYPE DESC;

• 오답

• 실행 결과