코딩테스트 연습(Level.2) / MySQL

코드 입력

SELECT 
    NAME,
    COUNT(ANIMAL_ID) AS 'COUNT'
FROM ANIMAL_INS
GROUP BY NAME
HAVING COUNT >= 2 AND NAME IS NOT NULL
ORDER BY NAME;

SELECT ANIMAL_ID,NAME,SEX_UPON_INTAKE from ANIMAL_INS
where name in ('Lucy', 'Ella', 'Pickle', 'Rogan', 'Sabrina', 'Mitty');

SELECT ANIMAL_ID, NAME FROM ANIMAL_INS
WHERE upper(NAME) LIKE '%EL%' AND ANIMAL_TYPE='Dog'
ORDER BY NAME ASC;

대문자 결과 확인했어야 함. upper 함수 참고

SELECT count(DISTINCT NAME) as count from ANIMAL_INS
where name is not null