database

https://school.programmers.co.kr/learn/courses/30/lessons/59045

SELECT O.ANIMAL_ID, O.ANIMAL_TYPE, O.NAME
FROM ANIMAL_INS as I
JOIN ANIMAL_OUTS as O ON I.ANIMAL_ID = O.ANIMAL_ID
WHERE I.SEX_UPON_INTAKE != O.SEX_UPON_OUTCOME

https://school.programmers.co.kr/learn/courses/30/lessons/62284
이 문제 살짝 절었다.

1번 풀이

WITH M AS (SELECT DISTINCT CART_ID, NAME
FROM CART_PRODUCTS
WHERE NAME = 'Milk')

, Y AS (SELECT DISTINCT CART_ID, NAME
FROM CART_PRODUCTS
WHERE NAME = 'Yogurt')

SELECT Y.CART_ID
FROM Y
INNER JOIN M
ON Y.CART_ID = M.CART_ID
ORDER BY Y.CART_ID

2번 풀이

SELECT CART_ID 
FROM CART_PRODUCTS
WHERE NAME IN ('Yogurt', 'Milk')
GROUP BY CART_ID
HAVING COUNT(DISTINCT NAME) >= 2

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그룹화 에러 발생 해결 방법
sql_mode=only_full_group_by 에러 해결 방법

https://info-lab.tistory.com/274