sjmin

문제 설명 예시 해결 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;

문제 설명 예시 해결 SELECT ANIMAL_ID, NAME, CASE WHEN SEX_UPON_INTAKE IN ('Neutered Male', 'Spayed Female') THEN 'O' ELSE 'X' END '중성화' FROM ANIMAL_INS;

문제 설명 예시 해결 SELECT ANIMAL_ID, NAME FROM ANIMAL_INS WHERE NAME LIKE '%el%' AND ANIMAL_TYPE = 'Dog' ORDER BY NAME;

문제 설명 예시 해결 SELECT ANIMAL_ID, NAME, SEX_UPON_INTAKE FROM ANIMAL_INS WHERE NAME IN ('Lucy', 'Ella', 'Pickle', 'Rogan', 'Sabrina', 'Mitty') ORDER BY ANIMAL_ID;

문제 설명 문제 예시 해결 SELECT HISTORY_ID, CAR_ID, DATE_FORMAT(START_DATE, '%Y-%m-%d') START_DATE, DATE_FORMAT(END_DATE, '%Y-%m-%d') END_DATE, CASE WHEN DATEDIFF(END_DATE, START_DATE) + 1 >= 30 THEN '장기 대여' ELSE '단기 대여' END RENT_TYPE FROM CAR_RENTAL_COMPANY_RENTAL_HISTORY WHERE DATE_FORMAT(START_DATE, '%Y-%m') = '2022-09' ORDER BY HISTORY_ID DESC;

문제 설명 문제 예시

문제 설명 문제 예시 해결 SELECT A.PRODUCT_CODE, SUM(A.PRICE*B.SALES_AMOUNT) SALE FROM PRODUCT A JOIN OFFLINE_SALE B ON A.PRODUCT_ID = B.PRODUCT_ID GROUP BY A.PRODUCT_ID ORDER BY 2 DESC, 1;

문제 설명 예시 해결 SELECT A.ANIMAL_ID, A.ANIMAL_TYPE, A.NAME FROM ANIMAL_INS A LEFT JOIN ANIMAL_OUTS B ON A.ANIMAL_ID = B.ANIMAL_ID WHERE A.SEX_UPON_INTAKE != B.SEX_UPON_OUTCOME;

문제 설명 예시 해결 SELECT NAME, DATETIME FROM ANIMAL_INS WHERE ANIMAL_ID NOT IN (SELECT ANIMAL_ID FROM ANIMAL_OUTS) ORDER BY DATETIME LIMIT 3; SELECT A.NAME, A.DATETIME FROM ANIMAL_INS A LEFT JOIN ANIMAL_OUTS B ON A.ANIMAL_ID = B.ANIMAL_ID WHERE B.ANIMAL_ID IS NULL ORDER BY A.DATETIME LIMIT 3;

문제 설명 예시 해결 SELECT B.ANIMAL_ID, B.NAME FROM ANIMAL_INS A INNER JOIN ANIMAL_OUTS B ON A.ANIMAL_ID = B.ANIMAL_ID WHERE A.DATETIME > B.DATETIME ORDER BY A.DATETIME;