SQL 고득점 Kit - SELECT Level4 - 서울에 위치한 식당 목록 출력하기 - mysql

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

select rest_info.rest_id, rest_name, food_type, favorites, address, avg_review_score as score
from rest_info
inner join(
    select rest_id as avg_rest_id, round(avg(review_score), 2) as avg_review_score
    from rest_review
    group by rest_id
) avg_rest_review on rest_info.rest_id = avg_rest_review.avg_rest_id
where address like '서울%'
order by score desc, favorites desc

round(숫자, 반올림 소수점 자릿수)

avg(<컬럼>)

inner join()