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[level 4] Title: 그룹별 조건에 맞는 식당 목록 출력하기, Time: 0.00 ms, Memory: 0.0 MB -BaekjoonHub
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‎프로그래머스/4/131124. 그룹별 조건에 맞는 식당 목록 출력하기/README.md‎

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# [level 4] 그룹별 조건에 맞는 식당 목록 출력하기 - 131124
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[문제 링크](https://school.programmers.co.kr/learn/courses/30/lessons/131124)
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### 성능 요약
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메모리: 0.0 MB, 시간: 0.00 ms
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### 구분
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코딩테스트 연습 > JOIN
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### 채점결과
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합계: 100.0 / 100.0
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### 제출 일자
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2026년 04월 09일 22:35:35
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### 문제 설명
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<p>다음은 고객의 정보를 담은 <code>MEMBER_PROFILE</code>테이블과 식당의 리뷰 정보를 담은 <code>REST_REVIEW</code> 테이블입니다. <code>MEMBER_PROFILE</code> 테이블은 다음과 같으며 <code>MEMBER_ID</code>, <code>MEMBER_NAME</code>, <code>TLNO</code>, <code>GENDER</code>, <code>DATE_OF_BIRTH</code>는 회원 ID, 회원 이름, 회원 연락처, 성별, 생년월일을 의미합니다.</p>
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<table class="table">
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<thead><tr>
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<th>Column name</th>
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<th>Type</th>
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<th>Nullable</th>
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</tr>
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</thead>
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<tbody><tr>
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<td>MEMBER_ID</td>
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<td>VARCHAR(100)</td>
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<td>FALSE</td>
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</tr>
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<tr>
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<td>MEMBER_NAME</td>
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<td>VARCHAR(50)</td>
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<td>FALSE</td>
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</tr>
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<tr>
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<td>TLNO</td>
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<td>VARCHAR(50)</td>
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<td>TRUE</td>
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</tr>
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<tr>
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<td>GENDER</td>
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<td>VARCHAR(1)</td>
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<td>TRUE</td>
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</tr>
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<tr>
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<td>DATE_OF_BIRTH</td>
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<td>DATE</td>
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<td>TRUE</td>
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</tr>
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</tbody>
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</table>
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<p><code>REST_REVIEW</code> 테이블은 다음과 같으며 <code>REVIEW_ID</code>, <code>REST_ID</code>, <code>MEMBER_ID</code>, <code>REVIEW_SCORE</code>, <code>REVIEW_TEXT</code>,<code>REVIEW_DATE</code>는 각각 리뷰 ID, 식당 ID, 회원 ID, 점수, 리뷰 텍스트, 리뷰 작성일을 의미합니다.</p>
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<table class="table">
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<thead><tr>
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<th>Column name</th>
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<th>Type</th>
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<th>Nullable</th>
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</tr>
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</thead>
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<tbody><tr>
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<td>REVIEW_ID</td>
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<td>VARCHAR(10)</td>
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<td>FALSE</td>
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</tr>
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<tr>
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<td>REST_ID</td>
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<td>VARCHAR(10)</td>
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<td>TRUE</td>
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</tr>
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<tr>
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<td>MEMBER_ID</td>
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<td>VARCHAR(100)</td>
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<td>TRUE</td>
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</tr>
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<tr>
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<td>REVIEW_SCORE</td>
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<td>NUMBER</td>
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<td>TRUE</td>
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</tr>
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<tr>
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<td>REVIEW_TEXT</td>
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<td>VARCHAR(1000)</td>
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<td>TRUE</td>
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</tr>
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<tr>
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<td>REVIEW_DATE</td>
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<td>DATE</td>
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<td>TRUE</td>
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</tr>
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</tbody>
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</table>
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<hr>
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<h5>문제</h5>
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<p><code>MEMBER_PROFILE</code>와 <code>REST_REVIEW</code> 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.</p>
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<hr>
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<h5>예시</h5>
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<p><code>MEMBER_PROFILE</code> 테이블이 다음과 같고</p>
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<table class="table">
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<thead><tr>
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<th>MEMBER_ID</th>
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<th>MEMBER_NAME</th>
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<th>TLNO</th>
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<th>GENDER</th>
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<th>DATE_OF_BIRTH</th>
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</tr>
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</thead>
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<tbody><tr>
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<td><code>jiho92@naver.com</code></td>
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<td>이지호</td>
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<td>01076432111</td>
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<td>W</td>
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<td>1992-02-12</td>
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</tr>
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<tr>
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<td><code>jiyoon22@hotmail.com</code></td>
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<td>김지윤</td>
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<td>01032324117</td>
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<td>W</td>
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<td>1992-02-22</td>
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</tr>
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<tr>
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<td><code>jihoon93@hanmail.net</code></td>
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<td>김지훈</td>
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<td>01023258688</td>
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<td>M</td>
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<td>1993-02-23</td>
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</tr>
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<tr>
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<td><code>seoyeons@naver.com</code></td>
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<td>박서연</td>
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<td>01076482209</td>
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<td>W</td>
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<td>1993-03-16</td>
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</tr>
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<tr>
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<td><code>yelin1130@gmail.com</code></td>
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<td>조예린</td>
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<td>01017626711</td>
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<td>W</td>
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<td>1990-11-30</td>
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</tr>
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</tbody>
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</table>
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<p><code>REST_REVIEW</code> 테이블이 다음과 같을 때</p>
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<table class="table">
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<thead><tr>
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<th>REVIEW_ID</th>
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<th>REST_ID</th>
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<th>MEMBER_ID</th>
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<th>REVIEW_SCORE</th>
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<th>REVIEW_TEXT</th>
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<th>REVIEW_DATE</th>
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</tr>
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</thead>
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<tbody><tr>
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<td>R000000065</td>
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<td>00028</td>
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<td><code>soobin97@naver.com</code></td>
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<td>5</td>
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<td>부찌 국물에서 샤브샤브 맛이나고 깔끔</td>
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<td>2022-04-12</td>
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</tr>
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<tr>
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<td>R000000066</td>
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<td>00039</td>
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<td><code>yelin1130@gmail.com</code></td>
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<td>5</td>
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<td>김치찌개 최곱니다.</td>
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<td>2022-02-12</td>
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</tr>
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<tr>
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<td>R000000067</td>
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<td>00028</td>
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<td><code>yelin1130@gmail.com</code></td>
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<td>5</td>
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<td>햄이 많아서 좋아요</td>
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<td>2022-02-22</td>
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</tr>
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<tr>
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<td>R000000068</td>
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<td>00035</td>
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<td><code>ksyi0316@gmail.com</code></td>
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<td>5</td>
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<td>숙성회가 끝내줍니다.</td>
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<td>2022-02-15</td>
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</tr>
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<tr>
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<td>R000000069</td>
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<td>00035</td>
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<td><code>yoonsy95@naver.com</code></td>
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<td>4</td>
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<td>비린내가 전혀없어요.</td>
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<td>2022-04-16</td>
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</tr>
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</tbody>
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</table>
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<p>SQL을 실행하면 다음과 같이 출력되어야 합니다.</p>
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<table class="table">
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<thead><tr>
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<th>MEMBER_NAME</th>
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<th>REVIEW_TEXT</th>
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<th>REVIEW_DATE</th>
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</tr>
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</thead>
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<tbody><tr>
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<td>조예린</td>
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<td>김치찌개 최곱니다.</td>
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<td>2022-02-12</td>
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</tr>
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<tr>
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<td>조예린</td>
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<td>햄이 많아서 좋아요</td>
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<td>2022-02-22</td>
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</tr>
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</tbody>
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</table>
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<hr>
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<h5>주의사항</h5>
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<p><code>REVIEW_DATE</code>의 데이트 포맷이 예시와 동일해야 정답처리 됩니다.</p>
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> 출처: 프로그래머스 코딩 테스트 연습, https://school.programmers.co.kr/learn/challenges

‎프로그래머스/4/131124. 그룹별 조건에 맞는 식당 목록 출력하기/그룹별 조건에 맞는 식당 목록 출력하기.sql‎

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-- 코드를 입력하세요
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SELECT MEMBER_NAME,
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REVIEW_TEXT,
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DATE_FORMAT(REVIEW_DATE, "%Y-%m-%d") AS REVIEW_DATE
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FROM MEMBER_PROFILE AS P
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JOIN REST_REVIEW AS R
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ON P.MEMBER_ID = R.MEMBER_ID
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WHERE R.MEMBER_ID IN (
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SELECT MEMBER_ID
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FROM REST_REVIEW
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GROUP BY MEMBER_ID
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HAVING COUNT(*) = (
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SELECT COUNT(*)
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FROM REST_REVIEW
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GROUP BY MEMBER_ID
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ORDER BY COUNT(*) DESC
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LIMIT 1
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)
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)
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ORDER BY REVIEW_DATE ASC, REVIEW_TEXT ASC;
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