[SQL] 코딩 테스트 연습 (3)

[SQL] 코딩 테스트 연습 (3)

https://school.programmers.co.kr/learn/challenges?tab=sql_practice_kit 

프로그래머스

 

1. FOOD_PRODUCT 테이블에서 식품분류별로 가격이 제일 비싼 식품의 분류, 가격, 이름을 조회하는 SQL문을 작성해주세요. 이때 식품분류가 '과자', '국', '김치', '식용유'인 경우만 출력시켜 주시고 결과는 식품 가격을 기준으로 내림차순 정렬해주세요.

SELECT CATEGORY, PRICE, PRODUCT_NAME
FROM FOOD_PRODUCT
WHERE 1=1
    AND CATEGORY IN ('과자','국','김치','식용유')
    AND PRICE IN (SELECT MAX(PRICE) 
                  FROM FOOD_PRODUCT 
                  GROUP BY CATEGORY)
GROUP BY CATEGORY
ORDER BY PRICE DESC

😵처음에는 아래와같이  코드를 사용했는데 GROUP BY로 묶으면 가장 상단에 있는 데이터들을 임의로 가져오기 때문에,

GROUP BY로 커버하는 것이 아닌 서브쿼리를 사용해야 함

SELECT CATEGORY, MAX(PRICE) AS MAX_PRICE, PRODUCT_NAME
FROM FOOD_PRODUCT
GROUP BY CATEGORY
HAVING CATEGORY IN ('과자', '국', '김치', '식용유')
ORDER BY MAX_PRICE DESC

 

2. CAR_RENTAL_COMPANY_CAR 테이블에서 '통풍시트', '열선시트', '가죽시트' 중 하나 이상의 옵션이 포함된 자동차가 자동차 종류 별로 몇 대인지 출력하는 SQL문을 작성해주세요. 이때 자동차 수에 대한 컬럼명은 CARS로 지정하고, 결과는 자동차 종류를 기준으로 오름차순 정렬해주세요.

SELECT CAR_TYPE, COUNT(CAR_ID) AS CARS
FROM CAR_RENTAL_COMPANY_CAR
WHERE 1=1
    AND (OPTIONS LIKE ('%통풍시트%') OR
         OPTIONS LIKE ('%열선시트%') OR 
         OPTIONS LIKE ('%가죽시트%'))
GROUP BY CAR_TYPE
ORDER BY CAR_TYPE

 

3. REST_INFO 테이블에서 음식종류별로 즐겨찾기수가 가장 많은 식당의 음식 종류, ID, 식당 이름, 즐겨찾기수를 조회하는 SQL문을 작성해주세요. 이때 결과는 음식 종류를 기준으로 내림차순 정렬해주세요.

SELECT FOOD_TYPE, REST_ID, REST_NAME, FAVORITES
FROM REST_INFO
WHERE 1=1
    AND FAVORITES IN (SELECT MAX(FAVORITES) 
                      FROM REST_INFO 
                      GROUP BY FOOD_TYPE)
GROUP BY FOOD_TYPE
ORDER BY FOOD_TYPE DESC

 

4. USED_GOODS_BOARD와 USED_GOODS_USER 테이블에서 완료된 중고 거래의 총금액이 70만 원 이상인 사람의 회원 ID, 닉네임, 총거래금액을 조회하는 SQL문을 작성해주세요. 결과는 총거래금액을 기준으로 오름차순 정렬해주세요.

SELECT B.USER_ID, B.NICKNAME, SUM(A.PRICE) AS TOTAL_PRICE
FROM USED_GOODS_BOARD A
    JOIN USED_GOODS_USER B
    ON A.WRITER_ID = B.USER_ID
WHERE STATUS = 'DONE'
GROUP BY USER_ID
HAVING TOTAL_PRICE >= 700000
ORDER BY TOTAL_PRICE

 

5. 2022년 1월의 카테고리 별 도서 판매량을 합산하고, 카테고리(CATEGORY), 총 판매량(TOTAL_SALES) 리스트를 출력하는 SQL문을 작성해주세요. 결과는 카테고리명을 기준으로 오름차순 정렬해주세요.

-- 코드를 입력하세요
SELECT A.CATEGORY, SUM(B.SALES) AS TOTAL_SALES
FROM BOOK A
    JOIN BOOK_SALES B
    ON A.BOOK_ID = B.BOOK_ID
WHERE B.SALES_DATE LIKE ('%2022-01%')
GROUP BY CATEGORY
ORDER BY CATEGORY

 

6. CAR_RENTAL_COMPANY_RENTAL_HISTORY 테이블에서 2022년 10월 16일에 대여 중인 자동차인 경우 '대여중' 이라고 표시하고, 대여 중이지 않은 자동차인 경우 '대여 가능'을 표시하는 컬럼(컬럼명: AVAILABILITY)을 추가하여 자동차 ID와 AVAILABILITY 리스트를 출력하는 SQL문을 작성해주세요. 이때 반납 날짜가 2022년 10월 16일인 경우에도 '대여중'으로 표시해주시고 결과는 자동차 ID를 기준으로 내림차순 정렬해주세요.

SELECT CAR_ID,
    CASE 
        WHEN MAX('2022-10-16' BETWEEN START_DATE AND END_DATE) THEN '대여중'
        ELSE '대여 가능'
    END AS AVAILABILITY
FROM CAR_RENTAL_COMPANY_RENTAL_HISTORY
GROUP BY CAR_ID
ORDER BY CAR_ID DESC

 

7.CAR_RENTAL_COMPANY_RENTAL_HISTORY 테이블에서 대여 시작일을 기준으로 2022년 8월부터 2022년 10월까지 총 대여 횟수가 5회 이상인 자동차들에 대해서 해당 기간 동안의 월별 자동차 ID 별 총 대여 횟수(컬럼명: RECORDS) 리스트를 출력하는 SQL문을 작성해주세요. 결과는 월을 기준으로 오름차순 정렬하고, 월이 같다면 자동차 ID를 기준으로 내림차순 정렬해주세요. 특정 월의 총 대여 횟수가 0인 경우에는 결과에서 제외해주세요.

SELECT MONTH(START_DATE) AS MONTH, CAR_ID, COUNT(*) AS RECORD
FROM CAR_RENTAL_COMPANY_RENTAL_HISTORY
WHERE 1=1
    AND CAR_ID IN (SELECT DISTINCT CAR_ID
                   FROM CAR_RENTAL_COMPANY_RENTAL_HISTORY
                   WHERE START_DATE BETWEEN '2022-08-01' AND '2022-10-31'
                   GROUP BY CAR_ID
                   HAVING COUNT(HISTORY_ID) >= 5) 
    AND START_DATE BETWEEN '2022-08-01' AND '2022-10-31'
GROUP BY MONTH, CAR_ID
ORDER BY MONTH, CAR_ID DESC

 

8. 보호소에서는 몇 시에 입양이 가장 활발하게 일어나는지 알아보려 합니다. 09:00부터 19:59까지, 각 시간대별로 입양이 몇 건이나 발생했는지 조회하는 SQL문을 작성해주세요. 이때 결과는 시간대 순으로 정렬해야 합니다.

SELECT HOUR(DATETIME) AS HOUR, COUNT(ANIMAL_ID) AS COUNT
FROM ANIMAL_OUTS
WHERE HOUR(DATETIME) BETWEEN 9 AND 20
GROUP BY HOUR(DATETIME)
ORDER by HOUR

 

9. USED_GOODS_BOARD와 USED_GOODS_USER 테이블에서 중고 거래 게시물을 3건 이상 등록한 사용자의 사용자 ID, 닉네임, 전체주소, 전화번호를 조회하는 SQL문을 작성해주세요. 이때, 전체 주소는 시, 도로명 주소, 상세 주소가 함께 출력되도록 해주시고, 전화번호의 경우 xxx-xxxx-xxxx 같은 형태로 하이픈 문자열(-)을 삽입하여 출력해주세요. 결과는 회원 ID를 기준으로 내림차순 정렬해주세요.

SELECT DISTINCT USER_ID, 
       NICKNAME,
       CONCAT(CITY, " ", STREET_ADDRESS1, " ", STREET_ADDRESS2) AS '전체주소', 
       CONCAT(SUBSTR(TLNO,1,3),'-',SUBSTR(TLNO,4,4),'-',SUBSTR(TLNO,8,11)) AS '전화번호'
FROM USED_GOODS_BOARD A
    INNER JOIN USED_GOODS_USER B
    ON A.WRITER_ID = B.USER_ID
WHERE USER_ID IN (SELECT WRITER_ID
                  FROM USED_GOODS_BOARD 
                  GROUP BY WRITER_ID
                  HAVING COUNT(BOARD_ID) >= 3)
ORDER BY USER_ID DESC;

🔎CONCAT

문자열을 결합하는데 사용되는 함수, 두 개 이상의 문자열을 하나로 결합 할 수 있음

CONCAT(string1, string2, ...)

예를 들어 아래 예시는 'Hello world!' 를 반환함

CONCAT('Hello', ' ', 'world!')

SUBSTR 함수와 같이 사용해서 문자열을 나눈뒤에 결합할 수 있음

 

10. USED_GOODS_BOARD와 USED_GOODS_FILE 테이블에서 조회수가 가장 높은 중고거래 게시물에 대한 첨부파일 경로를 조회하는 SQL문을 작성해주세요. 첨부파일 경로는 FILE ID를 기준으로 내림차순 정렬해주세요. 기본적인 파일경로는 /home/grep/src/ 이며, 게시글 ID를 기준으로 디렉토리가 구분되고, 파일이름은 파일 ID, 파일 이름, 파일 확장자로 구성되도록 출력해주세요. 조회수가 가장 높은 게시물은 하나만 존재합니다.

SELECT CONCAT("/home/grep/src/",A.BOARD_ID,"/",FILE_ID,FILE_NAME,FILE_EXT) AS FILE_PATH
FROM USED_GOODS_BOARD A
    JOIN USED_GOODS_FILE B
    ON A.BOARD_ID = B.BOARD_ID
WHERE A.BOARD_ID = (SELECT BOARD_ID 
                     FROM USED_GOODS_BOARD 
                     ORDER BY VIEWS DESC 
                     LIMIT 1)
ORDER BY FILE_ID DESC;

 

11. CAR_RENTAL_COMPANY_CAR 테이블과 CAR_RENTAL_COMPANY_RENTAL_HISTORY 테이블과 CAR_RENTAL_COMPANY_DISCOUNT_PLAN 테이블에서 자동차 종류가 '트럭'인 자동차의 대여 기록에 대해서 대여 기록 별로 대여 금액(컬럼명: FEE)을 구하여 대여 기록 ID와 대여 금액 리스트를 출력하는 SQL문을 작성해주세요. 결과는 대여 금액을 기준으로 내림차순 정렬하고, 대여 금액이 같은 경우 대여 기록 ID를 기준으로 내림차순 정렬해주세요.

WITH CTE AS (
    SELECT DISCOUNT_RATE, DURATION_TYPE
    FROM CAR_RENTAL_COMPANY_DISCOUNT_PLAN
    WHERE CAR_TYPE LIKE '트럭'
    )

SELECT HISTORY_ID, ROUND(DUR*DAILY_FEE*(100-RATE)/100) FEE
FROM
    (SELECT HISTORY_ID, DATEDIFF(END_DATE, START_DATE)+1 AS DUR,
        CASE
            WHEN DATEDIFF(END_DATE, START_DATE)+1 BETWEEN 7 AND 20 THEN (SELECT DISCOUNT_RATE FROM CTE WHERE DURATION_TYPE LIKE '7%')
            WHEN DATEDIFF(END_DATE, START_DATE)+1 BETWEEN 30 AND 89 THEN (SELECT DISCOUNT_RATE FROM CTE WHERE DURATION_TYPE LIKE '30%')
            WHEN DATEDIFF(END_DATE, START_DATE)+1 >= 90 THEN (SELECT DISCOUNT_RATE FROM CTE WHERE DURATION_TYPE LIKE '90%')
        ELSE 0
        END RATE, DAILY_FEE
    FROM CAR_RENTAL_COMPANY_RENTAL_HISTORY CC
        JOIN CAR_RENTAL_COMPANY_CAR CR 
        ON CR.CAR_ID = CC.CAR_ID
    WHERE CR.CAR_TYPE LIKE '트럭'
    ) SUB
ORDER BY FEE DESC, HISTORY_ID DESC

 

 

끝 😋