[프로그래머스] JOIN

joon_1592·2022년 4월 26일
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SQL

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SELECT ANIMAL_OUTS.ANIMAL_ID, ANIMAL_OUTS.NAME
FROM ANIMAL_INS RIGHT JOIN ANIMAL_OUTS 
ON ANIMAL_INS.ANIMAL_ID = ANIMAL_OUTS.ANIMAL_ID
WHERE ANIMAL_INS.ANIMAL_ID IS NULL
ORDER BY ANIMAL_INS.ANIMAL_ID

있었는데요 없었습니다

SELECT A_IN.ANIMAL_ID, A_IN.NAME
FROM ANIMAL_INS A_IN INNER JOIN ANIMAL_OUTS A_OUT
ON A_IN.ANIMAL_ID = A_OUT.ANIMAL_ID
WHERE A_IN.DATETIME > A_OUT.DATETIME
ORDER BY A_IN.DATETIME

오랜 기간 보호한 동물 (1)

SELECT A_IN.NAME, A_IN.DATETIME
FROM ANIMAL_INS A_IN LEFT JOIN ANIMAL_OUTS A_OUT
ON A_IN.ANIMAL_ID = A_OUT.ANIMAL_ID
WHERE A_OUT.ANIMAL_ID IS NULL
ORDER BY A_IN.DATETIME
LIMIT 3

보호소에서 중성화한 동물

SELECT A_OUT.ANIMAL_ID, A_OUT.ANIMAL_TYPE, A_OUT.NAME
FROM ANIMAL_INS A_IN INNER JOIN ANIMAL_OUTS A_OUT
ON A_IN.ANIMAL_ID = A_OUT.ANIMAL_ID
WHERE A_IN.SEX_UPON_INTAKE IN ('Intact Male', 'Intact Female')
AND A_OUT.SEX_UPON_OUTCOME IN ('Spayed Female', 'Neutered Male')
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