sjmin

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;

문제 설명 예시 해결 SELECT B.ANIMAL_ID, B.NAME FROM ANIMAL_INS A RIGHT JOIN ANIMAL_OUTS B ON A.ANIMAL_ID = B.ANIMAL_ID WHERE A.DATETIME IS NULL;

문제 설명 문제 예시 해결 SELECT FIRST_HALF.FLAVOR FROM FIRST_HALF ,(SELECT FLAVOR, SUM(TOTAL_ORDER) AS TOTAL FROM JULY GROUP BY FLAVOR) AS TMP WHERE TMP.FLAVOR = FIRST_HALF.FLAVOR ORDER BY (TOTAL + TOTAL_ORDER) DESC LIMIT 3;

문제 설명 문제 예시 해결 SELECT COUNT(*) USERS FROM USER_INFO WHERE AGE IS NULL;

문제 설명 예시 해결 SELECT ANIMAL_TYPE, IFNULL(NAME, 'No name') NAME, SEX_UPON_INTAKE FROM ANIMAL_INS;

문제 설명 예시 해결 SELECT ANIMAL_ID FROM ANIMAL_INS WHERE NAME IS NOT NULL;

문제 설명 예시 해결 SELECT ANIMAL_ID FROM ANIMAL_INS WHERE NAME IS NULL;