[프로그래머스] [MySQL] [Level 3] [JOIN] 있었는데요 없었습니다

---

---

-- 코드를 입력하세요
SELECT ANIMAL_INS.ANIMAL_ID, ANIMAL_INS.NAME
FROM ANIMAL_INS
LEFT JOIN ANIMAL_OUTS
ON ANIMAL_INS.ANIMAL_ID = ANIMAL_OUTS.ANIMAL_ID
WHERE ANIMAL_INS.DATETIME > ANIMAL_OUTS.DATETIME
ORDER BY ANIMAL_INS.DATETIME

SELECT A.ANIMAL_ID, A.NAME
FROM ANIMAL_INS AS A, ANIMAL_OUTS AS B 
WHERE A.ANIMAL_ID = B.ANIMAL_ID AND A.DATETIME > B.DATETIME
ORDER BY A.DATETIME;