[프로그래머스] [MySQL] [Level 3] [String, Date] 오랜 기간 보호한 동물(2)

---

---

SELECT A.ANIMAL_ID, A.NAME 
FROM ANIMAL_INS AS A, ANIMAL_OUTS AS B 
WHERE A.ANIMAL_ID = B.ANIMAL_ID 
ORDER BY B.DATETIME - A.DATETIME 
DESC LIMIT 2

SELECT ANIMAL_INS.ANIMAL_ID, ANIMAL_INS.NAME
FROM ANIMAL_INS
LEFT JOIN ANIMAL_OUTS
ON ANIMAL_INS.ANIMAL_ID = ANIMAL_OUTS.ANIMAL_ID
ORDER BY DATEDIFF(ANIMAL_OUTS.DATETIME, ANIMAL_INS.DATETIME) DESC
LIMIT 2;