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15 changes: 15 additions & 0 deletions sql/task1.sql
Original file line number Diff line number Diff line change
@@ -1,14 +1,29 @@
-- Problem 1: Retrieve all products in the Sports category
-- Write an SQL query to retrieve all products in a specific category.
SELECT p.product_id, p.product_name, p.price FROM Products p JOIN Categories c ON p.category_id = c.category_id WHERE c.category_name = 'Sports';

-- Problem 2: Retrieve the total number of orders for each user
-- Write an SQL query to retrieve the total number of orders for each user.
-- The result should include the user ID, username, and the total number of orders.
SELECT u.user_id, u.username, COUNT(o.order_id) AS total_orders
FROM Users u
JOIN Orders o ON u.user_id = o.user_id
GROUP BY u.user_id, u.username;

-- Problem 3: Retrieve the average rating for each product
-- Write an SQL query to retrieve the average rating for each product.
-- The result should include the product ID, product name, and the average rating.
SELECT p.product_id, p.product_name, AVG(r.rating) AS average_rating
FROM Products p
JOIN Reviews r ON p.product_id = r.product_id
GROUP BY p.product_id, p.product_name;

-- Problem 4: Retrieve the top 5 users with the highest total amount spent on orders
-- Write an SQL query to retrieve the top 5 users with the highest total amount spent on orders.
-- The result should include the user ID, username, and the total amount spent.
SELECT u.user_id, u.username, SUM(o.total_amount) AS total_amount_spent
FROM Users u
JOIN Orders o ON u.user_id = o.user_id
GROUP BY u.user_id, u.username
ORDER BY total_amount_spent DESC
LIMIT 5;
39 changes: 38 additions & 1 deletion sql/task2.sql
Original file line number Diff line number Diff line change
Expand Up @@ -2,18 +2,55 @@
-- Write an SQL query to retrieve the products with the highest average rating.
-- The result should include the product ID, product name, and the average rating.
-- Hint: You may need to use subqueries or common table expressions (CTEs) to solve this problem.
WITH avg_rating AS (
SELECT product_id, AVG(rating) AS average_rating
FROM Reviews
GROUP BY product_id;
)
SELECT p.product_id, p.product_name, ar.average_rating
FROM Products p
JOIN avg_rating ar ON p.product_id = ar.product_id
ORDER BY ar.average_rating DESC
LIMIT 1;

-- Problem 6: Retrieve the users who have made at least one order in each category
-- Write an SQL query to retrieve the users who have made at least one order in each category.
-- The result should include the user ID and username.
-- Hint: You may need to use subqueries or joins to solve this problem.
SELECT u.user_id, u.username
FROM Users u
WHERE NOT EXISTS (
SELECT c.category_id
FROM Categories c
WHERE NOT EXISTS (
SELECT o.order_id
FROM Orders o
JOIN Order_Items oi ON o.order_id = oi.order_id
JOIN Products p ON oi.product_id = p.product_id
WHERE o.user_id = u.user_id AND p.category_id = c.category_id
)
);


-- Problem 7: Retrieve the products that have not received any reviews
-- Write an SQL query to retrieve the products that have not received any reviews.
-- The result should include the product ID and product name.
-- Hint: You may need to use subqueries or left joins to solve this problem.
SELECT p.product_id, p.product_name
FROM Products p
LEFT JOIN Reviews r ON p.product_id = r.product_id
WHERE r.product_id IS NULL;

-- Problem 8: Retrieve the users who have made consecutive orders on consecutive days
-- Write an SQL query to retrieve the users who have made consecutive orders on consecutive days.
-- The result should include the user ID and username.
-- Hint: You may need to use subqueries or window functions to solve this problem.
-- Hint: You may need to use subqueries or window functions to solve this problem.
WITH ConsecutiveOrders AS (
SELECT user_id, order_date,
LAG(order_date) OVER (PARTITION BY user_id ORDER BY order_date) AS prev_order_date
FROM Orders
)
SELECT DISTINCT u.user_id, u.username
FROM Users u
JOIN ConsecutiveOrders co ON u.user_id = co.user_id
WHERE DATEDIFF(co.order_date, co.prev_order_date) = 1;
49 changes: 49 additions & 0 deletions sql/task3.sql
Original file line number Diff line number Diff line change
Expand Up @@ -2,18 +2,67 @@
-- Write an SQL query to retrieve the top 3 categories with the highest total sales amount.
-- The result should include the category ID, category name, and the total sales amount.
-- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
WITH CategorySales AS (
SELECT c.category_id, c.category_name, SUM(oi.quantity * oi.unit_price) AS total_sales
FROM Categories c
JOIN Products p ON c.category_id = p.category_id
JOIN Order_Items oi ON p.product_id = oi.product_id
GROUP BY c.category_id, c.category_name
)
SELECT category_id, category_name, total_sales
FROM CategorySales
ORDER BY total_sales DESC
LIMIT 3;

-- Problem 10: Retrieve the users who have placed orders for all products in the Toys & Games
-- Write an SQL query to retrieve the users who have placed orders for all products in the Toys & Games
-- The result should include the user ID and username.
-- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
SELECT u.user_id, u.username
FROM Users u
WHERE NOT EXISTS (
SELECT p.product_id
FROM Products p
JOIN Categories c ON p.category_id = c.category_id
WHERE c.category_name = 'Toys & Games'
AND NOT EXISTS (
SELECT oi.order_id
FROM Orders o
JOIN Order_Items oi ON o.order_id = oi.order_id
WHERE o.user_id = u.user_id
AND oi.product_id = p.product_id
)
);

-- Problem 11: Retrieve the products that have the highest price within each category
-- Write an SQL query to retrieve the products that have the highest price within each category.
-- The result should include the product ID, product name, category ID, and price.
-- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
WITH MaxPrices AS (
SELECT category_id, MAX(price) AS max_price
FROM Products
GROUP BY category_id
)
SELECT p.product_id, p.product_name, p.category_id, p.price
FROM Products p
JOIN MaxPrices mp ON p.category_id = mp.category_id AND p.price = mp.max_price;

-- Problem 12: Retrieve the users who have placed orders on consecutive days for at least 3 days
-- Write an SQL query to retrieve the users who have placed orders on consecutive days for at least 3 days.
-- The result should include the user ID and username.
-- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
WITH ConsecutiveOrders AS (
SELECT user_id, order_date,
LAG(order_date, 1) OVER (PARTITION BY user_id ORDER BY order_date) AS prev_order_date,
LAG(order_date, 2) OVER (PARTITION BY user_id ORDER BY order_date) AS prev_prev_order_date
FROM Orders
),
ConsecutiveDays AS (
SELECT user_id
FROM ConsecutiveOrders
WHERE DATEDIFF(order_date, prev_order_date) = 1
AND DATEDIFF(prev_order_date, prev_prev_order_date) = 1
)
SELECT DISTINCT u.user_id, u.username
FROM Users u
JOIN ConsecutiveDays cd ON u.user_id = cd.user_id;