#99. [LeetCode / MySQL] 1070. Product Sales Analysis III

문제 입력 코드 SELECT product_id, year AS first_year, quantity, price FROM Sales WHERE (product_id, year) in ( SELECT product_id, MIN(year) FROM Sales GROUP BY product_id ) 코드 설명 #SELECT #MIN #GROUP BY #SUBQUERY 문제 출처 LeetCode - The World's Leading Online Programming Learning Platform Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared f..

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• · 2023. 9. 25.
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#95. [LeetCode / MySQL] 1174. Immediate Food Delivery II

문제 입력 코드 SELECT ROUND(AVG(order_date = customer_pref_delivery_date) * 100, 2) AS immediate_percentage FROM Delivery WHERE (customer_id, order_date) IN ( SELECT customer_id, MIN(order_date) FROM Delivery GROUP BY customer_id ) 코드 설명 #SELECT #ROUND #AVG #MIN #GROUP BY #SUBQUERY 문제 출처 Immediate Food Delivery II - LeetCode Can you solve this real interview question? Immediate Food Delivery II - Tabl..

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• · 2023. 9. 21.
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#87. [LeetCode / MySQL] 570. Managers with at Least 5 Direct Reports

문제 입력 코드 1. SELF JOIN SELECT e1.name FROM Employee AS e1 JOIN Employee AS e2 ON e1.id = e2.managerId GROUP BY e1.id HAVING count(e2.managerId) >= 5; 2. SUBQUERY SELECT name FROM Employee WHERE id IN ( SELECT managerId FROM Employee GROUP BY managerId HAVING COUNT(managerId) >= 5 ) 코드 설명 #SELECT #JOIN #SELF JOIN #GROUP BY #HAVING #COUNT #SUBQUERY #IN 문제 출처 Managers with at Least 5 Direct Reports ..

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• · 2023. 9. 13.
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#76. [LeetCode / MySQL] 584. Find Customer Referee

문제 입력 코드 1. SELECT name FROM Customer WHERE (referee_id IS NULL) OR (referee_id != 2); 2. SELECT name FROM Customer WHERE id NOT IN ( SELECT id FROM Customer WHERE referee_id = 2 ); 코드 설명 #SELECT #NULL #NOT IN #SUBQUERY 문제 출처 Find Customer Referee - LeetCode Can you solve this real interview question? Find Customer Referee - Table: Customer +-------------+---------+ | Column Name | Type | +-----..

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• · 2023. 9. 2.
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#69. JOIN : 특정 기간동안 대여 가능한 자동차들의 대여비용 구하기 | 프로그래머스 | SQL 고득점 Kit

문제 입력 코드 SELECT CAR.CAR_ID, CAR.CAR_TYPE, ROUND((CAR.DAILY_FEE * ((100 - DISCOUNT.DISCOUNT_RATE)/100) * 30)) AS FEE FROM CAR_RENTAL_COMPANY_CAR AS CAR JOIN CAR_RENTAL_COMPANY_RENTAL_HISTORY AS RENT ON CAR.CAR_ID = RENT.CAR_ID JOIN CAR_RENTAL_COMPANY_DISCOUNT_PLAN AS DISCOUNT ON CAR.CAR_TYPE = DISCOUNT.CAR_TYPE WHERE CAR.CAR_ID NOT IN( SELECT CAR_ID FROM CAR_RENTAL_COMPANY_RENTAL_HISTORY WHERE EN..

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• · 2023. 8. 28.
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#68. JOIN : 주문량이 많은 아이스크림들 조회하기 | 프로그래머스 | SQL 고득점 Kit

문제 입력 코드 1. JOIN - ON SELECT FH.FLAVOR FROM FIRST_HALF AS FH JOIN JULY AS JUL ON FH.FLAVOR = JUL.FLAVOR GROUP BY FH.FLAVOR ORDER BY SUM(FH.TOTAL_ORDER) + SUM(JUL.TOTAL_ORDER) DESC LIMIT 3; 2. JOIN - USING SELECT FLAVOR FROM FIRST_HALF AS A JOIN JULY AS B USING (FLAVOR) GROUP BY FLAVOR ORDER BY SUM(A.TOTAL_ORDER + B.TOTAL_ORDER) DESC LIMIT 3; → 조인할 두 컬럼의 이름이 같은 경우에는 ON 대신에 USING을 사용할 수 있음 3. SUBQ..

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• · 2023. 8. 28.
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