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SQL19

[프로그래머스]오랜 기간 보호한 동물(1) (MySQL) JOIN -- # SELECT AI.NAME, AI.DATETIME -- # FROM ANIMAL_INS AI LEFT JOIN ANIMAL_OUTS AO -- # ON AI.ANIMAL_ID = AO.ANIMAL_ID -- # WHERE AO.ANIMAL_ID IS NULL -- # ORDER BY AI.DATETIME LIMIT 3; SELECT AI.NAME, AI.DATETIME FROM ANIMAL_INS AI LEFT JOIN ANIMAL_OUTS AO ON AI.ANIMAL_ID = AO.ANIMAL_ID WHERE AO.ANIMAL_ID IS NULL ORDER BY AI.DATETIME LIMIT 3; 2024. 1. 8.
프로그래머스 조건에 맞는 사용자와 총 거래 금액 조회하기(MySQL) -- 코드를 입력하세요 # SELECT USER_ID, NICKNAME, SUM(PRICE) AS TOTAL_SALES # FROM USED_GOODS_BOARD UGB JOIN USED_GOODS_USER UGU # ON UGB.WRITER_ID = UGU.USER_ID # WHERE STATUS = 'DONE' # GROUP BY USER_ID # HAVING SUM(PRICE) >=700000 # ORDER BY TOTAL_SALES; SELECT USER_ID, NICKNAME, SUM(PRICE) AS TOTAL_SALES FROM USED_GOODS_BOARD AS UGB LEFT JOIN USED_GOODS_USER AS UGU ON (UGB.WRITER_ID = UGU.USER_ID) WH.. 2024. 1. 3.
프로그래머스 즐겨찾기가 가장많은 식당정보 출력하기(MySQL) -- 코드를 입력하세요 # SELECT FOOD_TYPE, REST_ID, REST_NAME,FAVORITES # FROM REST_INFO # WHERE (FOOD_TYPE, FAVORITES) IN (SELECT FOOD_TYPE, MAX(FAVORITES) # FROM REST_INFO # GROUP BY FOOD_TYPE) # ORDER BY FOOD_TYPE DESC; SELECT FOOD_TYPE, REST_ID, REST_NAME, FAVORITES FROM REST_INFO WHERE (FOOD_TYPE, FAVORITES) IN (SELECT FOOD_TYPE, MAX(FAVORITES) FROM REST_INFO GROUP BY FOOD_TYPE) ORDER BY FOOD_TYPE DE.. 2024. 1. 3.
프로그래머스 중복 제거하기(MySQL) -- 코드를 입력하세요 SELECT COUNT(DISTINCT NAME) AS COUNT FROM ANIMAL_INS WHERE NAME IS NOT NULL; WHERE 절이 굳이 필요하진 않다. 왜냐하면 DISTINCT함수 자체가 NULL을 제외시키기 때문이다. 단지, 이해를 돕기위해 작성하였다. 2024. 1. 3.

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