[프로그래머스] [MySQL] [Level 4] [SELECT] 서울에 위치한 식당 목록 출력하기

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SELECT A.REST_ID, A.REST_NAME, A.FOOD_TYPE, A.FAVORITES, A.ADDRESS, ROUND(AVG(REVIEW_SCORE),2) AS SCORE
FROM REST_INFO AS A
LEFT JOIN REST_REVIEW AS B
ON A.REST_ID = B.REST_ID 
WHERE A.ADDRESS LIKE '서울%' AND REVIEW_SCORE IS NOT NULL
GROUP BY A.REST_ID
ORDER BY SCORE DESC, A.FAVORITES DESC;