https://school.programmers.co.kr/learn/courses/30/lessons/59042
IN 테이블에는 없는데,
OUT 테이블에는 있는 동물 정보 출력
SELECT
outs.ANIMAL_ID,
outs.NAME
FROM
ANIMAL_INS ins
RIGHT JOIN ANIMAL_OUTS outs
ON ins.ANIMAL_ID=outs.ANIMAL_ID
WHERE 1=1
AND ins.ANIMAL_ID IS NULL
ORDER BY
outs.ANIMAL_ID
eazy