[SQL] 프로그래머스 [Level-4] 그룹별 조건에 맞는 식당 목록 출력하기

<MySQL>

SELECT A.MEMBER_NAME
     , B.REVIEW_TEXT
     , DATE_FORMAT(B.REVIEW_DATE, '%Y-%m-%d') AS REVIEW_DATE
  FROM MEMBER_PROFILE A
     , REST_REVIEW B
 WHERE A.MEMBER_ID = B.MEMBER_ID
   AND B.MEMBER_ID IN (
                        SELECT MEMBER_ID
                          FROM REST_REVIEW
                      GROUP BY MEMBER_ID
                        HAVING COUNT(MEMBER_ID) = (
                                                    SELECT COUNT(MEMBER_ID) AS CNT
                                                      FROM REST_REVIEW
                                                  GROUP BY MEMBER_ID
                                                  ORDER BY COUNT(MEMBER_ID) DESC
                                                     LIMIT 1                      
                                                  )
                      )
 ORDER BY B.REVIEW_DATE, B.REVIEW_TEXT

<Oracle>

SELECT A.MEMBER_NAME
     , B.REVIEW_TEXT
     , TO_CHAR(B.REVIEW_DATE, 'YYYY-MM-DD') AS REVIEW_DATE
  FROM MEMBER_PROFILE A
     , REST_REVIEW B
 WHERE A.MEMBER_ID = B.MEMBER_ID
   AND B.MEMBER_ID IN (
                        SELECT MEMBER_ID
                          FROM REST_REVIEW
                      GROUP BY MEMBER_ID
                        HAVING COUNT(MEMBER_ID) = (
                                                    SELECT MAX(COUNT(MEMBER_ID)) AS CNT
                                                      FROM REST_REVIEW
                                                  GROUP BY MEMBER_ID                 
                                                  )
                      )
 ORDER BY B.REVIEW_DATE, B.REVIEW_TEXT

프로그래머스 그룹별 조건에 맞는 식당 목록 출력하기 SQL