![[SQL문제풀기] 서울에 위치한 식당 목록 출력하기](https://image.inblog.dev?url=https%3A%2F%2Fsource.inblog.dev%2Ffeatured_image%2F2025-07-29T09%3A17%3A03.548Z-8e4744a5-78bd-4b09-ab52-bec2eda3ac4d&w=3840&q=75)
문제
MYSQL
내가 작성한 오답
: 서울이라는 문자열이 주소 어디에 있든지 포함되면 출력된다.
SELECT I.REST_ID, I.REST_NAME, I.FOOD_TYPE, I.FAVORITES, I.ADDRESS, ROUND(AVG(R.REVIEW_SCORE),2) SCORE
FROM REST_INFO I
JOIN REST_REVIEW R ON I.REST_ID = R.REST_ID
WHERE I.ADDRESS LIKE '%서울%'
GROUP BY I.REST_ID, I.REST_NAME, I.FOOD_TYPE, I.FAVORITES, I.ADDRESS
ORDER BY 6 DESC, 4 DESC;내가 작성한 정답
: substring으로 ‘서울’이라는 글자가 앞에 존재할 때만 결과가 출력되게 한다
SELECT I.REST_ID, I.REST_NAME, I.FOOD_TYPE, I.FAVORITES, I.ADDRESS, ROUND(AVG(R.REVIEW_SCORE),2) SCORE
FROM REST_INFO I
JOIN REST_REVIEW R ON I.REST_ID = R.REST_ID
WHERE SUBSTRING(I.ADDRESS,1,2) LIKE '서울'
GROUP BY I.REST_ID, I.REST_NAME, I.FOOD_TYPE, I.FAVORITES, I.ADDRESS
ORDER BY 6 DESC, 4 DESC;ORACLE
내가 작성한 정답
SELECT I.REST_ID, I.REST_NAME, I.FOOD_TYPE, I.FAVORITES, I.ADDRESS, ROUND(AVG(R.REVIEW_SCORE),2) SCORE
FROM REST_INFO I
JOIN REST_REVIEW R ON I.REST_ID = R.REST_ID
WHERE SUBSTR(I.ADDRESS,1,2) = '서울'
GROUP BY I.REST_ID, I.REST_NAME, I.FOOD_TYPE, I.FAVORITES, I.ADDRESS
ORDER BY 6 DESC, 4 DESC;Share article