- 루시와 엘라 찾기 : too easy : in 문을 쓰는 문제인듯
select ANIMAL_ID, NAME, SEX_UPON_INTAKE from ANIMAL_INS as ani_i
where 1=1
and ani_i.NAME in ('Lucy', 'Ella', 'Pickle', 'Rogan', 'Sabrina', 'Mitty')
SELECT ANIMAL_ID, NAME
FROM ANIMAL_INS
WHERE ANIMAL_TYPE = 'Dog'
**AND UPPER(NAME) LIKE UPPER('%el%')**
ORDER BY NAME
SELECT ANIMAL_ID, NAME
FROM ANIMAL_INS
WHERE ANIMAL_TYPE = 'Dog'
**AND regexp_like(UPPER(NAME), ('el'), 'i)**
-- 출처 : https://jhnyang.tistory.com/292
ORDER BY NAME
SELECT ANIMAL_ID, NAME, CASE
WHEN regexp_like(SEX_UPON_INTAKE, ('Neutered|Spayed'))
THEN 'O'
ELSE 'X'
END as 중성화 -- 한글써도 되나?
FROM ANIMAL_INS
WHERE 1=1
ORDER BY ANIMAL_ID
SELECT ani_in.ANIMAL_ID, ani_in.NAME
FROM ANIMAL_INS ani_in JOIN ANIMAL_OUTS ani_out
ON ani_in.ANIMAL_ID = ani_out.ANIMAL_ID
WHERE 1=1
AND ani_in.NAME is not null
ORDER BY TIMESTAMPDIFF(second, ani_in.DATETIME, ani_out.DATETIME) desc
LIMIT 2
select ANIMAL_ID, NAME, DATE_FORMAT(DATETIME,'%Y-%m-%d') FROM ANIMAL_INS
order by ANIMAL_ID asc