[Programmers] 오랜 기간 보호한 동물(2)

문제 링크

정답

-- ANIMAL_INS: ANIMAL_ID, ANIMAL_TYPE, DATETIME, INTAKE_CONDITION, NAME, SEX_UPON_INTAKE
-- ANIMAL_OUTS: ANIMAL_ID, ANIMAL_TYPE, DATETIME, NAME, SEX_UPON_OUTCOME

WITH ANIMAL_DURATION AS (
    SELECT 
        T1.ANIMAL_ID, 
        T1.NAME,
        DATEDIFF(T2.DATETIME, T1.DATETIME) AS DURATION
    FROM ANIMAL_INS AS T1
    INNER JOIN ANIMAL_OUTS AS T2
        ON T1.ANIMAL_ID = T2.ANIMAL_ID
    ORDER BY 3 DESC
)

SELECT ANIMAL_ID, NAME
FROM ANIMAL_DURATION
LIMIT 2;

✅ Point

• 입양 간 동물: ANIMAL_INS, ANIMAL_OUTS 에 동시에 ID가 존재하는 동물