프로그래머스 고득점 키트_ JOIN_MYSQL

2020-04-24 · 배운 것 · 2분 읽기

https://programmers.co.kr/learn/challenges?selected_part_id=17046

프로그래머스

1. 없어진 기록찾기

select aout.ANIMAL_ID, aout.NAME
from ANIMAL_OUTS as aout
left join ANIMAL_INS as ain
on aout.ANIMAL_ID = ain.ANIMAL_ID
where ain.ANIMAL_ID is null

2. 있었는데요 없었습니다.

select ain.ANIMAL_ID, ain.NAME
from ANIMAL_INS as ain
join ANIMAL_OUTS as aout
on ain.ANIMAL_ID = aout.ANIMAL_ID
where ain.DATETIME > aout.DATETIME
order by ain.DATETIME

3. 오랜 기간 보호한 동물

select ain.NAME, ain.DATETIME
from ANIMAL_INS as ain
left join ANIMAL_OUTS as aout
on ain.ANIMAL_ID = aout.ANIMAL_ID
where aout.NAME is null and ain.NAME is not null
order by ain.DATETIME
limit 3

4. 보호소에서 중성화한 동물

select ain.ANIMAL_ID, ain.ANIMAL_TYPE, ain.NAME
from ANIMAL_INS as ain
inner join ANIMAL_OUTS as aout
on ain.ANIMAL_ID = aout.ANIMAL_ID
where ain.SEX_UPON_INTAKE like "Intact%"
and aout.SEX_UPON_OUTCOME like "Neutered%"
or ain.SEX_UPON_INTAKE like "Intact%"
and aout.SEX_UPON_OUTCOME like "Spayed%"
order by ain.ANIMAL_ID