[Programmers] 오랜 기간 보호한 동물(2)
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