sjmin

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;

문제 설명 문제 예시 해결 SELECT WAREHOUSE_ID, WAREHOUSE_NAME, ADDRESS, IF(FREEZER_YN IS NOT NULL, FREEZER_YN, 'N') AS FREEZER_YN FROM FOOD_WAREHOUSE WHERE ADDRESS LIKE '%경기도%' ORDER BY WAREHOUSE_ID;

문제 설명 문제 예시 해결 SELECT (PRICE - (PRICE % 10000)) PRICE_GROUP, COUNT(*) AS PRODUCTS FROM PRODUCT GROUP BY PRICE_GROUP ORDER BY PRICE_GROUP;

문제 설명 예시 해결 WITH RECURSIVE CTE AS ( SELECT 0 AS NUM UNION ALL SELECT NUM+1 FROM CTE WHERE NUM < 23 ) SELECT CTE.NUM HOUR, IFNULL(A.입양횟수, 0) COUNT FROM CTE LEFT JOIN (SELECT HOUR(datetime) 시간대, COUNT(*) 입양횟수 FROM ANIMAL_OUTS GROUP BY 시간대 ORDER BY 시간대) A ON CTE.NUM = A.시간대;