[Programmers] 있었는데요 없었습니다
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
SELECT I.ANIMAL_ID, I.NAME
FROM ANIMAL_INS AS I
INNER JOIN ANIMAL_OUTS AS O
ON I.ANIMAL_ID = O.ANIMAL_ID
WHERE I.DATETIME > O.DATETIME
ORDER BY I.DATETIME;
✅ Point
- I , O 둘 다 정확히 일치하는 정보 =>
INNER JOIN
로 접근