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[level 4] Title: 서울에 위치한 식당 목록 출력하기, Time: 0.00 ms, Memory: 0.0 MB -BaekjoonHub
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‎프로그래머스/4/131118. 서울에 위치한 식당 목록 출력하기/README.md‎

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# [level 4] 서울에 위치한 식당 목록 출력하기 - 131118
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[문제 링크](https://school.programmers.co.kr/learn/courses/30/lessons/131118)
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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 > SELECT
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### 채점결과
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Empty
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### 제출 일자
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2025년 11월 03일 19:36:09
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### 문제 설명
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<p>다음은 식당의 정보를 담은 <code>REST_INFO</code> 테이블과 식당의 리뷰 정보를 담은 <code>REST_REVIEW</code> 테이블입니다. <code>REST_INFO</code> 테이블은 다음과 같으며 <code>REST_ID</code>, <code>REST_NAME</code>, <code>FOOD_TYPE</code>, <code>VIEWS</code>, <code>FAVORITES</code>, <code>PARKING_LOT</code>, <code>ADDRESS</code>, <code>TEL</code>은 식당 ID, 식당 이름, 음식 종류, 조회수, 즐겨찾기수, 주차장 유무, 주소, 전화번호를 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>REST_ID</td>
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<td>VARCHAR(5)</td>
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<td>FALSE</td>
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</tr>
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<tr>
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<td>REST_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>FOOD_TYPE</td>
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<td>VARCHAR(20)</td>
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<td>TRUE</td>
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</tr>
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<tr>
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<td>VIEWS</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>FAVORITES</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>PARKING_LOT</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>ADDRESS</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>TEL</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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</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>REST_INFO</code>와 <code>REST_REVIEW</code> 테이블에서 서울에 위치한 식당들의 식당 ID, 식당 이름, 음식 종류, 즐겨찾기수, 주소, 리뷰 평균 점수를 조회하는 SQL문을 작성해주세요. 이때 리뷰 평균점수는 소수점 세 번째 자리에서 반올림 해주시고 결과는 평균점수를 기준으로 내림차순 정렬해주시고, 평균점수가 같다면 즐겨찾기수를 기준으로 내림차순 정렬해주세요. </p>
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<hr>
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<h5>예시</h5>
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<p><code>REST_INFO</code> 테이블이 다음과 같고</p>
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<table class="table">
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<thead><tr>
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<th>REST_ID</th>
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<th>REST_NAME</th>
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<th>FOOD_TYPE</th>
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<th>VIEWS</th>
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<th>FAVORITES</th>
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<th>PARKING_LOT</th>
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<th>ADDRESS</th>
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<th>TEL</th>
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</tr>
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</thead>
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<tbody><tr>
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<td>00028</td>
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<td>대우부대찌개</td>
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<td>한식</td>
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<td>52310</td>
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<td>10</td>
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<td>N</td>
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<td>경기도 용인시 처인구 남사읍 처인성로 309</td>
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<td>031-235-1235</td>
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</tr>
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<tr>
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<td>00039</td>
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<td>광주식당</td>
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<td>한식</td>
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<td>23001</td>
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<td>20</td>
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<td>N</td>
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<td>경기도 부천시 산업로8번길 60</td>
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<td>031-235-6423</td>
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</tr>
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<tr>
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<td>00035</td>
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<td>삼촌식당</td>
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<td>일식</td>
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<td>532123</td>
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<td>80</td>
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<td>N</td>
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<td>서울특별시 강서구 가로공원로76가길</td>
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<td>02-135-1266</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>REST_ID</th>
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<th>REST_NAME</th>
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<th>FOOD_TYPE</th>
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<th>FAVORITES</th>
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<th>ADDRESS</th>
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<th>SCORE</th>
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</tr>
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</thead>
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<tbody><tr>
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<td>00035</td>
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<td>삼촌식당</td>
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<td>일식</td>
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<td>80</td>
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<td>서울특별시 강서구 가로공원로76가길</td>
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<td>4.50</td>
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</tr>
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</tbody>
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</table>
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> 출처: 프로그래머스 코딩 테스트 연습, https://school.programmers.co.kr/learn/challenges

‎프로그래머스/4/131118. 서울에 위치한 식당 목록 출력하기/서울에 위치한 식당 목록 출력하기.sql‎

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-- 코드를 입력하세요
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SELECT
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INFO.REST_ID,
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INFO.REST_NAME,
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INFO.FOOD_TYPE,
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INFO.FAVORITES,
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INFO.ADDRESS,
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ROUND(AVG(REVIEW.REVIEW_SCORE), 2) AS SCORE
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FROM REST_INFO AS INFO
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JOIN REST_REVIEW AS REVIEW
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ON INFO.REST_ID = REVIEW.REST_ID
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WHERE INFO.ADDRESS LIKE '서울%'
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GROUP BY INFO.REST_ID
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ORDER BY SCORE DESC, INFO.FAVORITES DESC;

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