PROGRAMMING PROGRAMMING

PROGRAMMERS/SQL

• [프로그래머스] [MySQL] [Level 3] [Dev-Matching: 웹 백엔드 개발자(상반기)] 헤비 유저가 소유한 장소

  PROGRAMMERS/SQL 2022. 10. 14. 00:05

  SELECT * FROM PLACES WHERE HOST_ID IN (SELECT HOST_ID FROM PLACES GROUP BY HOST_ID HAVING COUNT(HOST_ID)>=2) ORDER BY ID

  Read More
• [프로그래머스] [MySQL] [Level 3] [JOIN] 없어진 기록 찾기

  PROGRAMMERS/SQL 2022. 10. 13. 22:11

  -- 코드를 입력하세요 SELECT A.ANIMAL_ID, A.NAME FROM ANIMAL_OUTS A LEFT OUTER JOIN ANIMAL_INS B ON A.ANIMAL_ID = B.ANIMAL_ID WHERE B.ANIMAL_ID is NULL ORDER BY A.ANIMAL_ID SELECT ANIMAL_OUTS.ANIMAL_ID, ANIMAL_OUTS.NAME FROM ANIMAL_OUTS LEFT JOIN ANIMAL_INS ON ANIMAL_INS.ANIMAL_ID = ANIMAL_OUTS.ANIMAL_ID WHERE ANIMAL_INS.ANIMAL_ID IS NULL ORDER BY ANIMAL_INS.ANIMAL_ID;

  Read More
• [프로그래머스] [MySQL] [Level 3] [JOIN] 있었는데요 없었습니다

  PROGRAMMERS/SQL 2022. 10. 13. 22:01

  -- 코드를 입력하세요 SELECT ANIMAL_INS.ANIMAL_ID, ANIMAL_INS.NAME FROM ANIMAL_INS LEFT JOIN ANIMAL_OUTS ON ANIMAL_INS.ANIMAL_ID = ANIMAL_OUTS.ANIMAL_ID WHERE ANIMAL_INS.DATETIME > ANIMAL_OUTS.DATETIME ORDER BY ANIMAL_INS.DATETIME SELECT A.ANIMAL_ID, A.NAME FROM ANIMAL_INS AS A, ANIMAL_OUTS AS B WHERE A.ANIMAL_ID = B.ANIMAL_ID AND A.DATETIME > B.DATETIME ORDER BY A.DATETIME;

  Read More
• [프로그래머스] [MySQL] [Level 3] [String, Date] 오랜 기간 보호한 동물(2)

  PROGRAMMERS/SQL 2022. 10. 13. 21:54

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

  Read More
• [프로그래머스] [MySQL] [Level 3] [JOIN] 오랜 기간 보호한 동물(1)

  PROGRAMMERS/SQL 2022. 10. 13. 21:46

  -- 코드를 입력하세요 SELECT ANIMAL_INS.NAME, ANIMAL_INS.DATETIME FROM ANIMAL_INS LEFT JOIN ANIMAL_OUTS ON ANIMAL_INS.ANIMAL_ID = ANIMAL_OUTS.ANIMAL_ID WHERE ANIMAL_OUTS.ANIMAL_ID is NULL ORDER BY DATETIME LIMIT 3

  Read More
• [프로그래머스] [MySQL] [Level 2] [GROUP BY] 가격대 별 상품 개수 구하기

  PROGRAMMERS/SQL 2022. 10. 13. 21:15

  SELECT TRUNCATE(PRICE, -4) AS PRICE_GROUP, COUNT(*) AS PRODUCTS FROM PRODUCT GROUP BY PRICE_GROUP ORDER BY PRICE_GROUP ASC; TRUNCATE(숫자,버릴 자릿수) ROUND(숫자,반올림할 자릿수)

  Read More
• [프로그래머스] [MySQL] [Level 2] [String, Date] 카테고리 별 상품 개수 구하기

  PROGRAMMERS/SQL 2022. 10. 13. 18:00

  SELECT LEFT(PRODUCT_CODE, 2) AS CATEGORY, COUNT(*) AS PRODUCTS FROM PRODUCT GROUP BY LEFT(PRODUCT_CODE, 2) ORDER BY PRODUCT_CODE;

  Read More
• [프로그래머스] [MySQL] [Level 3] [String, Date] 조건별로 분류하여 주문상태 출력하기

  PROGRAMMERS/SQL 2022. 10. 13. 17:47

  SELECT ORDER_ID, PRODUCT_ID, DATE_FORMAT(OUT_DATE, '%Y-%m-%d') AS OUR_DATE, CASE WHEN OUT_DATE IS NULL THEN '출고미정' WHEN DATEDIFF(OUT_DATE, "2022-05-01") > 0 THEN '출고대기' ELSE '출고완료' END AS '출고여부' FROM FOOD_ORDER ORDER BY ORDER_ID ASC; DATEDIFF DATEDIFF(날짜1, 날짜2); 날짜1 - 날짜2 (날짜가 뒤일수록 큰 수) TIMESTAMPDIFF TIMESTAMPDIFF(단위, 날짜1, 날짜2); 날짜2 - 날짜1 두 날짜 간의 차이를 단위 로 표현 단위 SECOND - 초 MINUTE - 분 HOUR - 시 DAY..

  Read More