[프로그래머스] JOIN > 없어진 기록 찾기

-- MySQL SELECT ANIMAL_OUTS.ANIMAL_ID, ANIMAL_OUTS.NAME from ANIMAL_OUTS LEFT JOIN ANIMAL_INS ON ANIMAL_OUTS.ANIMAL_ID = ANIMAL_INS.ANIMAL_ID where ANIMAL_INS.ANIMAL_ID IS NULL order by ANIMAL_OUTS.ANIMAL_ID; -- Oracle SELECT A.ANIMAL_ID, A.NAME FROM ANIMAL_OUTS A LEFT JOIN ANIMAL_INS B ON A.ANIMAL_ID = B.ANIMAL_ID WHERE B.DATETIME IS NULL ORDER BY A.ANIMAL_ID ASC;
Problem Solution/Programmers 2022. 10. 6. 02:00
[프로그래머스] IS NULL > 이름이 없는 동물의 아이디

-- MySQL SELECT ANIMAL_ID from ANIMAL_INS where NAME IS NULL; -- Oracle SELECT ANIMAL_ID FROM ANIMAL_INS WHERE NAME IS NULL ORDER BY ANIMAL_ID ASC;
Problem Solution/Programmers 2022. 10. 6. 01:59
[프로그래머스] IS NULL > 이름이 있는 동물의 아이디

-- MySQL SELECT ANIMAL_ID from ANIMAL_INS where NAME IS NOT NULL; -- Oracle SELECT ANIMAL_ID FROM ANIMAL_INS WHERE NAME IS NOT NULL ORDER BY ANIMAL_ID ASC;
Problem Solution/Programmers 2022. 10. 6. 01:59
[프로그래머스] IS NULL > NULL 처리하기

-- MySQL SELECT ANIMAL_TYPE, IFNULL(NAME,"No name") AS NAME, SEX_UPON_INTAKE from ANIMAL_INS -- Oracle SELECT ANIMAL_TYPE, NVL(NAME,'No name'), SEX_UPON_INTAKE FROM ANIMAL_INS ORDER BY ANIMAL_ID ASC;
Problem Solution/Programmers 2022. 10. 6. 01:58
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