[programmers]SQL_JOIN
💯 solving-algo | March 17, 2021
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SELECT OUTS.ANIMAL_ID, OUTS.NAME
FROM ANIMAL_OUTS AS OUTS
LEFT OUTER JOIN ANIMAL_INS AS INS
ON OUTS.ANIMAL_ID = INS.ANIMAL_ID
WHERE INS.ANIMAL_ID IS NULL
ORDER BY ANIMAL_ID있었는데요 없었습니다
# -- 코드를 입력하세요
SELECT INS.ANIMAL_ID, INS.NAME
FROM ANIMAL_INS AS INS
LEFT JOIN ANIMAL_OUTS AS OUTS
ON INS.ANIMAL_ID = OUTS.ANIMAL_ID
WHERE INS.DATETIME > OUTS.DATETIME
ORDER BY INS.DATETIME
# SELECT ANIMAL_INS.ANIMAL_ID, ANIMAL_INS.NAME
# FROM ANIMAL_INS
# LEFT JOIN ANIMAL_OUTS
# ON ANIMAL_INS.ANIMAL_ID=ANIMAL_OUTS.ANIMAL_ID
# WHERE ANIMAL_INS.DATETIME>ANIMAL_OUTS.DATETIME
# ORDER BY ANIMAL_INS.DATETIME오랜 기간 보호한 동물(1)
-- 코드를 입력하세요
SELECT INS.NAME, INS.DATETIME
FROM ANIMAL_INS AS INS
LEFT JOIN ANIMAL_OUTS AS OUTS
ON INS.ANIMAL_ID = OUTS.ANIMAL_ID
WHERE OUTS.ANIMAL_ID IS NULL
ORDER BY INS.DATETIME
LIMIT 3보호소에서 중성화한 동물
-- 코드를 입력하세요
SELECT OUTS.ANIMAL_ID, OUTS.ANIMAL_TYPE, OUTS.NAME
FROM ANIMAL_OUTS AS OUTS
INNER JOIN ANIMAL_INS AS INS
ON OUTS.ANIMAL_ID = INS.ANIMAL_ID
WHERE OUTS.SEX_UPON_OUTCOME <> INS.SEX_UPON_INTAKE
ORDER BY ANIMAL_ID