프로그래머스 SQL 76문제 풀기 ( 정답률 높은 순 기준 20개 ) - 1

LV.2 최솟값 구하기 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59038

 

SELECT DATETIME as '시간'
FROM ANIMAL_INS
ORDER BY DATETIME ASC limit 1;

 

LV.2 중복 제거하기 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59408

SELECT count(distinct NAME) as 'count'
FROM ANIMAL_INS
WHERE NAME IS NOT NULL;

 

LV.2 동물 수 구하기 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59406

SELECT count(*)
FROM ANIMAL_INS;

 

LV.2 동명 동물 수 찾기 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59041

SELECT NAME, COUNT(NAME)
FROM ANIMAL_INS
GROUP BY NAME
HAVING COUNT(NAME) > 1
ORDER BY NAME;

 

 

LV.1 동물의 아이디와 이름 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59403

SELECT ANIMAL_ID, NAME
FROM ANIMAL_INS;

 

 

LV.1 여러 기준으로 정렬하기 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59404

SELECT ANIMAL_ID, NAME, DATETIME
FROM ANIMAL_INS
ORDER BY NAME ASC, DATETIME DESC;

 

 

LV.1 이름이 있는 동물의 아이디 (정답률 : 91%)

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

SELECT ANIMAL_ID
FROM ANIMAL_INS
WHERE NAME IS NOT NULL;

 

LV.1 역순 정렬하기 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59035

SELECT NAME, DATETIME
FROM ANIMAL_INS
ORDER BY ANIMAL_ID DESC;

 

LV.1 어린 동물 찾기 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59037

SELECT ANIMAL_ID, NAME
FROM ANIMAL_INS
WHERE INTAKE_CONDITION <> 'Aged'
ORDER BY ANIMAL_ID ASC;

 

LV.1 상위 n개 레코드 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59405

SELECT NAME
FROM ANIMAL_INS
ORDER BY DATETIME LIMIT 1;

 

LV.1 아픈 동물 찾기 (정답률 : 91%)

https://school.programmers.co.kr/learn/courses/30/lessons/59036

SELECT ANIMAL_ID, NAME
FROM ANIMAL_INS
WHERE INTAKE_CONDITION = 'Sick';

 

 

LV.2 이름에 el이 들어가는 동물 찾기 (정답률 : 91%)

https://school.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;

 

LV.1 나이 정보가 없는 회원 수 구하기 (정답률 : 91%)

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

SELECT COUNT(*) AS 'USERS'
FROM USER_INFO
WHERE AGE IS NULL;

 

LV.2 NULL처리하기 (정답률 : 90%)

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

SELECT ANIMAL_TYPE, IFNULL(NAME, 'No name') as NAME, SEX_UPON_INTAKE
FROM ANIMAL_INS;

 

 

LV.1 가장 비싼 상품 구하기 (정답률 : 90%)

https://school.programmers.co.kr/learn/courses/30/lessons/131697

SELECT PRICE AS MAX_PRICE
FROM PRODUCT
ORDER BY PRICE DESC LIMIT 1;

SELECT MAX(PRICE) AS MAX_PRICE 
FROM PRODUCT;

 

LV.1 경기도에 위치한 식품창고 목록 출력하기 (정답률 : 90%)

https://school.programmers.co.kr/learn/courses/30/lessons/131114

SELECT WAREHOUSE_ID, WAREHOUSE_NAME, ADDRESS, IFNULL(FREEZER_YN, 'N') AS FREEZER_YN
FROM FOOD_WAREHOUSE
WHERE WAREHOUSE_NAME LIKE "%경기%";

 

LV.2 DATETIME에서 DATE로 형변환 (정답률 : 90%)

https://school.programmers.co.kr/learn/courses/30/lessons/59414

SELECT ANIMAL_ID, NAME, DATE_FORMAT(DATETIME, '%Y-%m-%d')  AS '날짜'
FROM ANIMAL_INS;

 

LV.2 가격이 제일 비싼 식품의 정보 출력하기 (정답률 : 89%)

https://school.programmers.co.kr/learn/courses/30/lessons/131115

SELECT PRODUCT_ID, PRODUCT_NAME, PRODUCT_CD, CATEGORY, PRICE
FROM FOOD_PRODUCT 
ORDER BY PRICE DESC LIMIT 1;

SELECT * 
FROM FOOD_PRODUCT 
WHERE PRICE = 
(SELECT MAX(PRICE) PRICE FROM FOOD_PRODUCT);

 

LV.1 이름이 없는 동물의 아이디 (정답률 : 89%)

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

SELECT ANIMAL_ID
FROM ANIMAL_INS
WHERE NAME IS NULL;

 

LV.1 강원도에 위치한 생산공장 목록 출력하기 (정답률 : 89%)

https://school.programmers.co.kr/learn/courses/30/lessons/131112

SELECT FACTORY_ID, FACTORY_NAME, ADDRESS
FROM FOOD_FACTORY
WHERE ADDRESS LIKE "강원도%";