[PROGRAMMERS] SUM, MAX, MIN
[Notification]
DBMS : MySQL
1. SUM, MAX, MIN LV.1
1. 가장 비싼 상품 구하기
1.1 정답
SELECT MAX(PRICE) AS MAX_PRICE FROM PRODUCT
2. 잡은 물고기 중 가장 큰 물고기의 길이 구하기
2.1 정답
SELECT
CONCAT(LENGTH, "cm") AS MAX_LENGTH
FROM
FISH_INFO
ORDER BY
LENGTH DESC
LIMIT 1;
3. 최댓값 구하기
3.1 정답
SELECT
DATETIME AS "시간"
FROM
ANIMAL_INS
ORDER BY
DATETIME DESC
limit 1
2. SUM, MAX, MIN LV.2
1. 연도별 대장균 크기의 편차 구하기
1.1 정답
SELECT
A.YEAR,
(B.MAX_SIZE - A.SIZE_OF_COLONY) AS YEAR_DEV,
A.ID
FROM
(SELECT
YEAR(DIFFERENTIATION_DATE) AS YEAR,
SIZE_OF_COLONY,
ID
FROM
ECOLI_DATA) AS A
JOIN
(SELECT
YEAR(DIFFERENTIATION_DATE) AS YEAR,
MAX(SIZE_OF_COLONY) AS MAX_SIZE
FROM
ECOLI_DATA
GROUP BY
YEAR) AS B
ON
A.YEAR = B.YEAR
ORDER BY
YEAR ASC,
YEAR_DEV ASC;
2. 조건에 맞는 아이템들의 가격의 총합 구하기
2.1 정답
SELECT
SUM(PRICE) as TOTAL_PRICE
FROM
ITEM_INFO
WHERE
RARITY = "LEGEND"
3. 가격이 제일 비싼 식품의 정보 출력하기
3.1 정답
SELECT PR
ODUCT_ID,PRODUCT_NAME,PRODUCT_CD,CATEGORY,PRICE
FROM
FOOD_PRODUCT
WHERE
PRICE = (SELECT MAX(PRICE) FROM FOOD_PRODUCT)
4. 중복 제거하기
4.1 정답
SELECT
COUNT(DISTINCT NAME) as "count"
FROM
ANIMAL_INS
5. 동물 수 구하기
5.1 정답
SELECT
count(*) as "count"
FROM
ANIMAL_INS
6. 최솟값 구하기
6.1 정답
SELECT
MIN(DATETIME) AS "시간"
FROM
ANIMAL_INS
OR
SELECT
DATETIME AS "시간"
FROM
ANIMAL_INS
ORDER BY
DATETIME ASC
limit 1
3. SUM, MAX, MIN LV.3
1. 물고기 종류 별 대어 찾기
1.1 정답
SELECT
A.ID,
B.FISH_NAME,
A.MAX_LENGTH AS LENGTH
FROM
(
SELECT
B.ID,
A.FISH_TYPE,
A.MAX_LENGTH
FROM
(
SELECT
FISH_TYPE,
MAX(LENGTH) AS MAX_LENGTH
FROM
FISH_INFO
GROUP BY
FISH_TYPE
) AS A,
FISH_INFO AS B
WHERE
B.FISH_TYPE = A.FISH_TYPE
AND B.LENGTH = A.MAX_LENGTH
) AS A,
FISH_NAME_INFO AS B
WHERE
A.FISH_TYPE = B.FISH_TYPE
ORDER BY
A.ID ASC;
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