[SQL문제풀기] 보호소에서 중성화한 동물

문제

MYSQL

SELECT I.ANIMAL_ID, I.ANIMAL_TYPE, I.NAME
FROM ANIMAL_INS I
JOIN ANIMAL_OUTS O ON I.ANIMAL_ID = O.ANIMAL_ID
WHERE I.SEX_UPON_INTAKE LIKE 'Intact%'
    AND O.SEX_UPON_OUTCOME NOT LIKE 'Intact%'
ORDER BY I.ANIMAL_ID ASC;

ORACLE

내가 작성한 오답

SELECT I.ANIMAL_ID, I.ANIMAL_TYPE, I.NAME
FROM ANIMAL_INS I
JOIN ANIMAL_OUTS O ON I.ANIMAL_ID = O.ANIMAL_ID
WHERE I.SEX_UPON_INTAKE LIKE 'Intact%'
        AND O.SEX_UPON_OUTCOME NOT LIKE 'Intact%'
ORDER BY I.ANIMAL_ID ASC;

내가 작성한 정답

SELECT I.ANIMAL_ID, I.ANIMAL_TYPE, I.NAME
FROM ANIMAL_INS I
JOIN ANIMAL_OUTS O ON I.ANIMAL_ID = O.ANIMAL_ID
WHERE I.SEX_UPON_INTAKE LIKE '%Intact%'
        AND O.SEX_UPON_OUTCOME NOT LIKE '%Intact%'
ORDER BY I.ANIMAL_ID ASC