sjmin

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;

문제 설명 문제 예시 해결 SELECT YEAR(SALES_DATE) "YEAR", MONTH(SALES_DATE) MONTH, GENDER, COUNT(DISTINCT(B.USER_ID)) USERS FROM USER_INFO A, ONLINE_SALE B WHERE A.USER_ID = B.USER_ID AND A.GENDER IS NOT NULL GROUP BY MONTH, GENDER ORDER BY 1,2,3;

문제 설명 예시 해결 SELECT NAME, COUNT(NAME) COUNT FROM ANIMAL_INS GROUP BY NAME HAVING COUNT(NAME) > 1 ORDER BY NAME;

문제 설명 예시 해결 SELECT ANIMAL_TYPE, COUNT(ANIMAL_TYPE) COUNT FROM ANIMAL_INS GROUP BY ANIMAL_TYPE ORDER BY ANIMAL_TYPE;

문제 설명 문제 예시 해결 SELECT FOOD_TYPE, REST_ID, REST_NAME, FAVORITES FROM REST_INFO WHERE FAVORITES IN (SELECT MAX(FAVORITES) FROM REST_INFO GROUP BY FOOD_TYPE) GROUP BY FOOD_TYPE ORDER BY FOOD_TYPE DESC;

문제 설명 예시 해결 SELECT COUNT(DISTINCT(NAME)) COUNT FROM ANIMAL_INS;

문제 설명 예시 해결 SELECT COUNT(*) "count" FROM ANIMAL_INS;

문제 설명 예시 해결 SELECT MIN(DATETIME) "시간" FROM ANIMAL_INS;