프로그래머스/[프로그래머스 - ORACLE] Lv.4
[프로그래머스 - ORACLE] 서울에 위치한 식당 목록 출력하기
코딩하는 흰둥이
2023. 3. 23. 17:47
CREATE 문
CREATE TABLE REST_INFO (
REST_ID VARCHAR(5) NOT NULL,
REST_NAME VARCHAR(50) NOT NULL,
FOOD_TYPE VARCHAR(20) NULL,
VIEWS NUMBER NULL,
FAVORITES NUMBER NULL,
PARKING_LOT VARCHAR(1) NULL,
ADDRESS VARCHAR(100) NULL,
TEL VARCHAR(100) NULL
)
CREATE TABLE REST_REVIEW (
REVIEW_ID VARCHAR(10) NOT NULL,
REST_ID VARCHAR(10) NULL,
MEMBER_ID VARCHAR(100) NULL,
REVIEW_SCORE NUMBER NULL,
REVIEW_TEXT VARCHAR(1000) NULL,
REVIEW_DATE DATE NULL
)
- 내 풀이
SELECT
I.REST_ID,
I.REST_NAME,
I.FOOD_TYPE,
I.FAVORITES,
I.ADDRESS,
ROUND(AVG(R.REVIEW_SCORE),2) AS SCORE
FROM REST_INFO I INNER JOIN REST_REVIEW R
ON I.REST_ID = R.REST_ID
AND I.ADDRESS LIKE '서울%'
GROUP BY I.REST_ID, I.REST_NAME, I.FOOD_TYPE, I.FAVORITES, I.ADDRESS
ORDER BY SCORE DESC , FAVORITES DESC;
왜 4단계가 더 어려운거 같지...