<IS NULL>
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59039
->
SELECT ANIMAL_ID
FROM ANIMAL_INS
WHERE NAME IS NULL
ORDER BY ANIMAL_ID
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59407
->
SELECT ANIMAL_ID
FROM ANIMAL_INS
WHERE NAME IS NOT NULL
ORDER BY ANIMAL_ID
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59410
->
SELECT ANIMAL_TYPE, IFNULL(NAME,"No name"), SEX_UPON_INTAKE
FROM ANIMAL_INS
ORDER BY ANIMAL_ID
<JOIN>
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59042
->
SELECT A.ANIMAL_ID, A.NAME
FROM ANIMAL_OUTS AS A LEFT JOIN ANIMAL_INS AS B ON A.ANIMAL_ID = B.ANIMAL_ID
WHERE B.ANIMAL_ID IS NULL
ORDER BY A.ANIMAL_ID
->
SELECT ANIMAL_ID, NAME
FROM ANIMAL_OUTS
WHERE ANIMAL_ID NOT IN (SELECT ANIMAL_ID FROM ANIMAL_INS)
ORDER BY ANIMAL_ID
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59043
->
SELECT A.ANIMAL_ID, A.NAME
FROM ANIMAL_INS A LEFT JOIN ANIMAL_OUTS B ON A.ANIMAL_ID = B.ANIMAL_ID
WHERE A.DATETIME > B.DATETIME
ORDER BY A.DATETIME
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59044
->
SELECT A.NAME, A.DATETIME
FROM ANIMAL_INS A LEFT JOIN ANIMAL_OUTS B ON A.ANIMAL_ID = B. ANIMAL_ID
WHERE B.ANIMAL_ID IS NULL
ORDER BY A.DATETIME
LIMIT 3
문제 링크 :programmers.co.kr/learn/courses/30/lessons/59045
->
SELECT A.ANIMAL_ID, A.ANIMAL_TYPE, A.NAME
FROM ANIMAL_INS A LEFT JOIN ANIMAL_OUTS B ON A.ANIMAL_ID = B.ANIMAL_ID
WHERE B.SEX_UPON_OUTCOME NOT Like 'Intact%' AND A.SEX_UPON_INTAKE LIKE 'Intact%'
ORDER BY A.ANIMAL_ID
<String, Date>
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59046
->
SELECT ANIMAL_ID, NAME, SEX_UPON_INTAKE
FROM ANIMAL_INS
WHERE NAME = 'Lucy' OR NAME = 'Ella' OR NAME = 'Pickle' OR NAME = 'Rogan' OR NAME = 'Sabrina' OR NAME = 'Mitty'
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59047
->
SELECT ANIMAL_ID, NAME
FROM ANIMAL_INS
WHERE NAME Like '%el%' AND ANIMAL_TYPE = 'DOG'
ORDER BY NAME
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59409
->
SELECT ANIMAL_ID, NAME,
CASE
WHEN SEX_UPON_INTAKE LIKE 'Neutered%' OR SEX_UPON_INTAKE LIKE 'Spayed%'
THEN 'O'
ELSE
'X'
END AS 중성화
FROM ANIMAL_INS
ORDER BY ANIMAL_ID
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59411
->
SELECT B.ANIMAL_ID, B.NAME
FROM ANIMAL_INS A RIGHT JOIN ANIMAL_OUTS B ON A.ANIMAL_ID = B.ANIMAL_ID
ORDER BY B.DATETIME - A.DATETIME DESC
LIMIT 2
문제 링크 : programmers.co.kr/learn/courses/30/lessons/59414
->
SELECT ANIMAL_ID, NAME, DATE_FORMAT(DATETIME, '%Y-%m-%d') AS 날짜
FROM ANIMAL_INS
ORDER BY ANIMAL_ID
'CS(Computer Science) > 데이터 베이스' 카테고리의 다른 글
프로그래머스 SQL 문제 (MySQL) - 1 (0) | 2020.11.07 |
---|---|
데이터 베이스 2(Database) (0) | 2020.09.25 |
데이터 베이스 1(Database) (0) | 2020.09.24 |