[LeetCode] 1141. User Activity for the Past 30 Days I [문제] [MySQL]SELECT activity_date AS day , COUNT(DISTINCT user_id) AS active_usersFROM ActivityWHERE activity_date > DATE_SUB('2019-07-27', INTERVAL 30 DAY) AND activity_date SQL/LeetCode 2025.02.19
[LeetCode] 1084. Sales Analysis III [문제] [MySQL]SELECT p.product_id , p.product_nameFROM Product AS p LEFT JOIN Sales AS s ON p.product_id = s.product_idGROUP BY p.product_idHAVING MIN(s.sale_date) >= '2019-01-01' AND MAX(s.sale_date) SQL/LeetCode 2025.02.19
[LeetCode] 1075. Project Employees I [문제] [MySQL]WITH Project_employee AS ( SELECT p.* , e.experience_years FROM Project AS p LEFT JOIN Employee AS e ON p.employee_id = e.employee_id)SELECT project_id , ROUND(AVG(experience_years), 2) AS average_yearsFROM Project_employeeGROUP BY project_id SQL/LeetCode 2025.02.19
[LeetCode] 1070. Product Sales Analysis III [문제] [MySQL]WITH first_year AS ( SELECT product_id , MIN(year) AS year FROM Sales GROUP BY product_id)SELECT s.product_id , f.year AS first_year , s.quantity , s.priceFROM Sales AS s LEFT JOIN first_year AS f ON s.product_id = f.product_idWHERE s.year = f.year SQL/LeetCode 2025.02.19
[LeetCode] 1068. Product Sales Analysis I [문제] [MySQL]SELECT p.product_name , s.year , s.priceFROM Sales AS s LEFT JOIN Product AS p ON s.product_id = p.product_id SQL/LeetCode 2025.02.19
[LeetCode] 1050. Actors and Directors Who Cooperated At Least Three Times [문제] [MySQL]SELECT actor_id, director_idFROM ActorDirectorGROUP BY actor_id, director_idHAVING COUNT(*) >= 3 SQL/LeetCode 2025.02.19
[LeetCode] 1045. Customers Who Bought All Products [문제] [MySQL]SELECT customer_idFROM CustomerGROUP BY customer_idHAVING COUNT(DISTINCT product_key) = (SELECT COUNT(*) FROM Product) SQL/LeetCode 2025.02.19
[LeetCode] 627. Swap Salary [문제] [MySQL]UPDATE salarySET sex = (CASE WHEN sex = 'm' THEN 'f' ELSE 'm' END) SQL/LeetCode 2025.02.18
[LeetCode] 626. Exchange Seats [문제] [MySQL]WITH id_row AS ( SELECT * , LAG(student, 1) OVER (ORDER BY id ASC) AS front_student , LEAD(student, 1) OVER (ORDER BY id ASC) AS back_student FROM Seat)SELECT id , CASE WHEN (MOD(id, 2) = 1 AND back_student IS NULL) THEN student WHEN MOD(id, 2) = 1 THEN back_student WHEN MOD(id, 2) = 0 THEN front_student END AS studentFROM id_row SQL/LeetCode 2025.02.18
[LeetCode] 620. Not Boring Movies [문제] [MySQL]SELECT *FROM CinemaWHERE MOD(id, 2) = 1 AND description != 'boring'ORDER BY rating DESC SQL/LeetCode 2025.02.18
