sjmin

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;

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

문제 설명 문제 예시 해결 SELECT * FROM FOOD_PRODUCT WHERE PRICE = (SELECT MAX(PRICE) FROM FOOD_PRODUCT); SELECT * FROM FOOD_PRODUCT ORDER BY PRICE DESC LIMIT 1;

문제 설명 문제 예시 해결 SELECT COUNT(USER_ID) USERS FROM USER_INFO WHERE AGE BETWEEN 20 AND 29 AND JOINED LIKE '2021%'; SELECT count(USER_ID) USERS FROM USER_INFO WHERE YEAR(JOINED) = 2021 AND AGE BETWEEN 20 AND 29;