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💻 하나씩 차곡차곡/프로그래머스 (SQL)105

[GROUP BY/LEVEL4] 입양 시각 구하기(2) (MySQL/Oracle) MySQLSET @HOUR = -1;SELECT (@HOUR := @HOUR +1) AS HOUR , (SELECT COUNT(HOUR(DATETIME)) FROM ANIMAL_OUTS WHERE HOUR(DATETIME)=@HOUR) AS COUNT FROM ANIMAL_OUTSWHERE @HOUR   OracleSELECT D.HOUR, NVL(O.CNT,0)AS COUNT FROM (SELECT LEVEL - 1 AS HOUR FROM DUAL CONNECT BY LEVEL 2023. 4. 21.
[GROUP BY/LEVEL2] 입양 시각 구하기(1) (MySQL/Oracle) MySQLSELECT HOUR(DATETIME), COUNT(*) AS COUNTFROM ANIMAL_OUTSWHERE HOUR(DATETIME) >= 9 AND HOUR(DATETIME)   OracleSELECT CASE WHEN TO_CHAR(DATETIME,'HH24') LIKE '0%'THEN SUBSTR(TO_CHAR(DATETIME,'HH24'),2,2) ELSE TO_CHAR(DATETIME,'HH24') END AS HOUR , COUNT(DATETIME) FROM ANIMAL_OUTS GROUP BY TO_CHAR(DATETIME,'HH24') HAVING TO_CHAR(DATETIME,'HH24') BETWEEN '09' AND '19'ORDE.. 2023. 4. 20.
[GROUP BY/LEVEL2] 가격대 별 상품 개수 구하기(MySQL/Oracle) MySQL/OracleSELECT SUBSTR(PRICE, 1, 1) * 10000 AS PRICE_GROUP ,COUNT(*) AS PRODUCTSFROM PRODUCTGROUP BY SUBSTR(PRICE,1,1)*10000ORDER BY PRICE_GROUP 2023. 4. 19.
[SUM,MAX,MIN/LEVEL2] 가격이 제일 비싼 식품의 정보 출력하기 (MySQL/Oracle) MySQL/OracleSELECT PRODUCT_ID , PRODUCT_NAME , PRODUCT_CD , CATEGORY , PRICEFROM FOOD_PRODUCTWHERE PRICE IN ( SELECT MAX(PRICE) FROM FOOD_PRODUCT )또 다른 풀이 (MySQL)SELECT PRODUCT_ID , PRODUCT_NAME , PRODUCT_CD , CATEGORY , PRICEFROM FOOD_PRODUCTORDER BY PRICE DESCLIMIT 1.. 2023. 4. 18.
[SUM,MAX,MIN/LEVEL2] 최솟값 구하기 (MySQL/Oracle) MySQL/OracleSELECT MIN(DATETIME)FROM ANIMAL_INS 2023. 4. 17.
[SUM,MAX,MIN/LEVEL2] 동물 수 구하기 (MySQL/Oracle) MySQL/OracleSELECT COUNT(*)FROM ANIMAL_INS 2023. 4. 16.
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