[프로그래머스] IS NULL (MySQL, Oracle)

경기도에 위치한 식품창고 목록 출력하기(Lv.1)

https://school.programmers.co.kr/learn/courses/30/lessons/131114?language=mysql 

# MySQL
# IFNULL
SELECT WAREHOUSE_ID, WAREHOUSE_NAME, ADDRESS, IFNULL(FREEZER_YN, 'N') AS FREEZER_YN
FROM FOOD_WAREHOUSE
WHERE ADDRESS LIKE '경기도%'
ORDER BY WAREHOUSE_ID

# COALESCE
SELECT WAREHOUSE_ID, WAREHOUSE_NAME, ADDRESS, COALESCE(FREEZER_YN, "N") AS FREEZER_YN
FROM FOOD_WAREHOUSE
WHERE ADDRESS LIKE '경기도%'
ORDER BY WAREHOUSE_ID

# CASE-ELSE
SELECT WAREHOUSE_ID, WAREHOUSE_NAME, ADDRESS, 
    CASE WHEN FREEZER_YN IS NULL THEN 'N'
    ELSE FREEZER_YN END AS FREEZER_YN
FROM FOOD_WAREHOUSE
WHERE ADDRESS LIKE '경기도%'
ORDER BY WAREHOUSE_ID


# ORACLE
SELECT WAREHOUSE_ID, WAREHOUSE_NAME, ADDRESS, NVL(FREEZER_YN, 'N') AS FREEZER_YN
FROM FOOD_WAREHOUSE
WHERE ADDRESS LIKE '경기도%'
ORDER BY WAREHOUSE_ID

 

 

이름이 없는 동물의 ID(Lv.1)

https://school.programmers.co.kr/learn/courses/30/lessons/59039

프로그래머스

# MySQL, ORACLE
SELECT ANIMAL_ID 
FROM ANIMAL_INS 
WHERE NAME IS NULL
ORDER BY ANIMAL_ID

 

이름이 있는 동물의 ID(Lv.1)

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

프로그래머스

 

# MySQL, ORACLE
SELECT ANIMAL_ID 
FROM ANIMAL_INS 
WHERE NAME IS NOT NULL
ORDER BY ANIMAL_ID

 

 

NULL 처리하기 (Lv.2)

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

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

# ORACLE
SELECT ANIMAL_TYPE, NVL(NAME, 'No name') as NAME, SEX_UPON_INTAKE
FROM ANIMAL_INS  
ORDER BY ANIMAL_ID

 

나이 정보가 없는 회원 수 구하기 (Lv.2)

https://school.programmers.co.kr/learn/courses/30/lessons/131528

프로그래머스

# MySQL, ORACLE
SELECT count(*) AS USERS
FROM USER_INFO
WHERE AGE IS NULL