[SQL문제풀기] 오랜 기간 보호한 동물(2)

문제

MYSQL

내가 작성한 정답

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

ORACLE

내가 작성한 정답

SELECT I.ANIMAL_ID, I.NAME
FROM ANIMAL_INS I
JOIN ANIMAL_OUTS O ON I.ANIMAL_ID = O.ANIMAL_ID
ORDER BY O.DATETIME-I.DATETIME DESC
FETCH FIRST 2 ROWS ONLY;

SELECT ANIMAL_ID, NAME
FROM (SELECT I.ANIMAL_ID, I.NAME
      FROM ANIMAL_INS I
      JOIN ANIMAL_OUTS O ON I.ANIMAL_ID = O.ANIMAL_ID
      ORDER BY O.DATETIME - I.DATETIME DESC)
WHERE ROWNUM <=2;