31. 오랜 기간 보호한 동물(1)--작성한 코드 >> 오답SELECT ai.NAME, ai.DATETIMEFROM ANIMAL_INS AS ai LEFT JOIN ANIMAL_OUTS AS aoON ai.ANIMAL_ID = ao.ANIMAL_IDWHERE ai.ANIMAL_ID NOT IN (ao.ANIMAL_ID)ORDER BY ai.DATETIME;--정답1SELECT AI.NAME, AI.DATETIMEFROM ANIMAL_INS AI LEFT JOIN ANIMAL_OUTS AOON AI.ANIMAL_ID=AO.ANIMAL_IDWHERE AO.ANIMAL_ID IS NULL -- 추가 설명 참조ORDER BY AI.DATETIMELIMIT 3;--정답2SELECT NAME, DATETIMEFRO..
