[SQL] 프로그래머스 보호소에서 중성화한 동물

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

난이도

Level 4

문제

코드

MySQL

SELECT I.ANIMAL_ID, I.ANIMAL_TYPE, I.NAME
FROM ANIMAL_INS I LEFT JOIN ANIMAL_OUTS O
ON I.ANIMAL_ID = O.ANIMAL_ID
WHERE O.DATETIME IS NOT NULL 
    AND I.SEX_UPON_INTAKE LIKE 'Intact%'
    AND (O.SEX_UPON_OUTCOME LIKE 'Neutered%'
         OR O.SEX_UPON_OUTCOME LIKE 'Spayed%')
ORDER BY ANIMAL_ID;

Oracle

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 LIKE 'Neutered%'
    OR O.SEX_UPON_OUTCOME LIKE 'Spayed%')
ORDER BY 1;