SQL 고득점 Kit - GROUP BY Level3 - 즐겨찾기가 가장 많은 식당 정보 출력하기- mysql

https://school.programmers.co.kr/learn/challenges?page=1&languages=mysql&order=recent

select food_type, rest_id, rest_name, favorites
from rest_info
inner join (
    select food_type as food_type1, max(favorites) as max_fav
    from rest_info
    group by food_type    
) join1 on rest_info.food_type = join1.food_type1 and rest_info.favorites = join1.max_fav
order by food_type desc

inner join key 값 2개