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출처: 프로그래머스

https://programmers.co.kr/learn/courses/30/lessons/59411

SELECT A.ANIMAL_ID, A.NAME
FROM ANIMAL_INS AS A
INNER JOIN ANIMAL_OUTS AS B
ON A.ANIMAL_ID = B.ANIMAL_ID
ORDER BY B.DATETIME-A.DATETIME DESC
LIMIT 2;

* ORDER BY 절에도 연산자 가능

SELECT A.ANIMAL_ID, A.NAME
FROM ANIMAL_INS AS A
INNER JOIN ANIMAL_OUTS AS B
ON A.ANIMAL_ID = B.ANIMAL_ID
ORDER BY DATEDIFF(B.DATETIME,A.DATETIME) DESC
LIMIT 2;

* 날짜 차이 함수 DATEDIFF(a,b)

출처: 프로그래머스

https://programmers.co.kr/learn/courses/30/lessons/59409

SELECT ANIMAL_ID, 
    NAME, 
    CASE
        WHEN SEX_UPON_INTAKE LIKE '%Neutered%' THEN 'O'
        WHEN SEX_UPON_INTAKE LIKE '%Spayed%' THEN 'O'
        ELSE 'X'
    END AS 중성화
FROM ANIMAL_INS
ORDER BY ANIMAL_ID;

출처: 프로그래머스 

https://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;

* LIKE 는 대소문자 구별을 하지 않는다.

출처: 프로그래머스

https://programmers.co.kr/learn/courses/30/lessons/59046

-- 코드를 입력하세요
SELECT ANIMAL_ID, 
    NAME,
    SEX_UPON_INTAKE
FROM ANIMAL_INS
WHERE NAME IN ('Lucy', 'Ella', 'Pickle', 'Rogan', 'Sabrina', 'Mitty')
ORDER BY ANIMAL_ID;

출처: 프로그래머스

https://programmers.co.kr/learn/courses/30/lessons/59410

1. IF 문

SELECT ANIMAL_TYPE, 
	IF(NAME IS NULL,'No name',NAME) AS NAME,
	SEX_UPON_INTAKE
FROM ANIMAL_INS
ORDER BY ANIMAL_ID;

2. IFNULL() 

SELECT ANIMAL_TYPE, 
	IFNULL(NAME,'No name') AS NAME,
	SEX_UPON_INTAKE
FROM ANIMAL_INS
ORDER BY ANIMAL_ID;

3. CASE 문

SELECT ANIMAL_TYPE, 
	CASE
    	WHEN NAME IS NULL THEN 'No name'
    	ELSE NAME
	END AS NAME,
	SEX_UPON_INTAKE
FROM ANIMAL_INS
ORDER BY ANIMAL_ID;

출처: 프로그래머스

https://programmers.co.kr/learn/courses/30/lessons/59407

-- 코드를 입력하세요
SELECT ANIMAL_ID
FROM ANIMAL_INS
WHERE NAME IS NOT NULL
ORDER BY ANIMAL_ID;

출처: 프로그래머스

-- 코드를 입력하세요
SELECT ANIMAL_ID
FROM ANIMAL_INS
WHERE NAME IS NULL;