[programmers] SQL Lv. 4 - 그룹별 조건에 맞는 식당 목록 출력하기

https://school.programmers.co.kr/learn/courses/30/lessons/131124

프로그래머스

문제 설명
다음은 고객의 정보를 담은 MEMBER_PROFILE테이블과 식당의 리뷰 정보를 담은 REST_REVIEW 테이블입니다. MEMBER_PROFILE 테이블은 다음과 같으며 MEMBER_ID, MEMBER_NAME, TLNO, GENDER, DATE_OF_BIRTH는 회원 ID, 회원 이름, 회원 연락처, 성별, 생년월일을 의미합니다.

REST_REVIEW 테이블은 다음과 같으며 REVIEW_ID, REST_ID, MEMBER_ID, REVIEW_SCORE, REVIEW_TEXT,REVIEW_DATE는 각각 리뷰 ID, 식당 ID, 회원 ID, 점수, 리뷰 텍스트, 리뷰 작성일을 의미합니다.

 

문제
MEMBER_PROFILE와 REST_REVIEW 테이블에서 리뷰를 가장 많이 작성한 회원의 리뷰들을 조회하는 SQL문을 작성해주세요. 회원 이름, 리뷰 텍스트, 리뷰 작성일이 출력되도록 작성해주시고, 결과는 리뷰 작성일을 기준으로 오름차순, 리뷰 작성일이 같다면 리뷰 텍스트를 기준으로 오름차순 정렬해주세요.

 

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풀이

회원 별 리뷰 작성 개수를 출력하는 쿼리

SELECT MEMBER_ID, COUNT(*) AS REVIEW_COUNT
FROM REST_REVIEW
GROUP BY MEMBER_ID

 

MAX 리뷰 개수를 출력하는 쿼리

SELECT MAX(REVIEW_COUNT)
FROM (
    SELECT MEMBER_ID, COUNT(*) AS REVIEW_COUNT
    FROM REST_REVIEW
    GROUP BY MEMBER_ID
) SUB

 

최다 리뷰 작성 회원을 출력하는 쿼리

-- 코드를 입력하세요
SELECT *
FROM REST_REVIEW R
JOIN MEMBER_PROFILE M
    ON M.MEMBER_ID = R.MEMBER_ID
WHERE R.MEMBER_ID IN (
    SELECT MEMBER_ID
    FROM REST_REVIEW
    GROUP BY MEMBER_ID
    HAVING COUNT(*) = (
        SELECT MAX(REVIEW_COUNT)
        FROM (
            SELECT MEMBER_ID, COUNT(*) AS REVIEW_COUNT
            FROM REST_REVIEW
            GROUP BY MEMBER_ID
        ) SUB
    )
)

• `HAVING`절에서 MAX 리뷰 개수와 동일한 리뷰 개수를 가진 회원만 필터링

 

문제 조건에 맞는 컬럼, DATE_FORMAT, ORDER BY 지정

SELECT M.MEMBER_NAME, 
    R.REVIEW_TEXT, 
    DATE_FORMAT(R.REVIEW_DATE, '%Y-%m-%d') AS REVIEW_DATE
FROM REST_REVIEW R
JOIN MEMBER_PROFILE M
    ON M.MEMBER_ID = R.MEMBER_ID
WHERE R.MEMBER_ID IN (
    SELECT MEMBER_ID
    FROM REST_REVIEW
    GROUP BY MEMBER_ID
    HAVING COUNT(*) = (
        SELECT MAX(REVIEW_COUNT)
        FROM (
            SELECT MEMBER_ID, COUNT(*) AS REVIEW_COUNT
            FROM REST_REVIEW
            GROUP BY MEMBER_ID
        ) SUB
    )
)
ORDER BY R.REVIEW_DATE, R.REVIEW_TEXT

계속 틀려서 뭔가 했더니 DATE_FORMAT이 따로 있었다..

... 문제를 잘 읽자

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제출

-- 코드를 입력하세요
SELECT M.MEMBER_NAME, 
    R.REVIEW_TEXT, 
    DATE_FORMAT(R.REVIEW_DATE, '%Y-%m-%d') AS REVIEW_DATE
FROM REST_REVIEW R
JOIN MEMBER_PROFILE M
    ON M.MEMBER_ID = R.MEMBER_ID
WHERE R.MEMBER_ID IN (
    SELECT MEMBER_ID
    FROM REST_REVIEW
    GROUP BY MEMBER_ID
    HAVING COUNT(*) = (
        SELECT MAX(REVIEW_COUNT)
        FROM (
            SELECT MEMBER_ID, COUNT(*) AS REVIEW_COUNT
            FROM REST_REVIEW
            GROUP BY MEMBER_ID
        ) SUB
    )
)
ORDER BY R.REVIEW_DATE, R.REVIEW_TEXT