Finished assesment

sql/task1.sql

@@ -12,3 +12,31 @@
 -- 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 * 
+FROM products
+WHERE category = 'Sports';
+
+SELECT users.user_id, users.username, COUNT(orders.order_id) AS total_orders
+FROM users
+LEFT JOIN orders ON users.user_id = orders.user_id
+GROUP BY users.user_id, users.username;
+
+SELECT products.product_id, products.product_name, AVG(ratings.rating) AS average_rating
+FROM products
+LEFT JOIN ratings ON products.product_id = ratings.product_id
+GROUP BY products.product_id, products.product_name;
+
+SELECT users.user_id, users.username, SUM(order_details.total_price) AS total_amount_spent
+FROM users
+LEFT JOIN (
+    SELECT orders.user_id, order_details.product_id, SUM(order_details.quantity * products.price) AS total_price
+    FROM orders
+    LEFT JOIN order_details ON orders.order_id = order_details.order_id
+    LEFT JOIN products ON order_details.product_id = products.product_id
+    GROUP BY orders.user_id, order_details.product_id
+) AS order_details ON users.user_id = order_details.user_id
+GROUP BY users.user_id, users.username
+ORDER BY total_amount_spent DESC
+LIMIT 5;
+

sql/task2.sql

@@ -16,4 +16,46 @@
 -- 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 AvgProductRatings AS (
+    SELECT products.product_id, products.product_name, AVG(ratings.rating) AS average_rating
+    FROM products
+    LEFT JOIN ratings ON products.product_id = ratings.product_id
+    GROUP BY products.product_id, products.product_name
+)
+SELECT product_id, product_name, average_rating
+FROM AvgProductRatings
+WHERE average_rating = (
+    SELECT MAX(average_rating) FROM AvgProductRatings
+);
+
+
+SELECT users.user_id, users.username
+FROM users
+WHERE users.user_id IN (
+    SELECT DISTINCT users.user_id
+    FROM users
+    JOIN orders ON users.user_id = orders.user_id
+    JOIN order_details ON orders.order_id = order_details.order_id
+    JOIN products ON order_details.product_id = products.product_id
+    GROUP BY users.user_id
+    HAVING COUNT(DISTINCT products.category) = (
+        SELECT COUNT(DISTINCT category) FROM products
+    )
+);
+
+SELECT products.product_id, products.product_name
+FROM products
+LEFT JOIN ratings ON products.product_id = ratings.product_id
+WHERE ratings.rating IS NULL;
+
+WITH UserOrdersWithDates AS (
+    SELECT orders.user_id, orders.order_date, LAG(orders.order_date) OVER (PARTITION BY orders.user_id ORDER BY orders.order_date) AS prev_order_date
+    FROM orders
+)
+SELECT DISTINCT user_id, username
+FROM users
+JOIN UserOrdersWithDates ON users.user_id = UserOrdersWithDates.user_id
+WHERE DATEDIFF(order_date, prev_order_date) = 1;
+

sql/task3.sql

@@ -17,3 +17,60 @@
 -- 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 CategorySales AS (
+    SELECT products.category_id, SUM(order_details.quantity * products.price) AS total_sales
+    FROM order_details
+    JOIN products ON order_details.product_id = products.product_id
+    GROUP BY products.category_id
+)
+SELECT categories.category_id, categories.category_name, COALESCE(total_sales, 0) AS total_sales
+FROM categories
+LEFT JOIN CategorySales ON categories.category_id = CategorySales.category_id
+ORDER BY total_sales DESC
+LIMIT 3;
+
+SELECT users.user_id, users.username
+FROM users
+WHERE users.user_id IN (
+    SELECT DISTINCT users.user_id
+    FROM users
+    JOIN orders ON users.user_id = orders.user_id
+    JOIN order_details ON orders.order_id = order_details.order_id
+    JOIN products ON order_details.product_id = products.product_id
+    WHERE products.category = 'Toys & Games'
+    GROUP BY users.user_id
+    HAVING COUNT(DISTINCT products.product_id) = (
+        SELECT COUNT(DISTINCT product_id) FROM products WHERE category = 'Toys & Games'
+    )
+);
+
+WITH MaxPricePerCategory AS (
+    SELECT products.product_id, products.product_name, products.category_id, products.price,
+           ROW_NUMBER() OVER(PARTITION BY products.category_id ORDER BY products.price DESC) AS rn
+    FROM products
+)
+SELECT product_id, product_name, category_id, price
+FROM MaxPricePerCategory
+WHERE rn = 1;
+
+WITH UserOrdersWithDates AS (
+    SELECT orders.user_id, orders.order_date, LAG(orders.order_date) OVER (PARTITION BY orders.user_id ORDER BY orders.order_date) AS prev_order_date
+    FROM orders
+)
+SELECT DISTINCT user_id, username
+FROM users
+JOIN UserOrdersWithDates ON users.user_id = UserOrdersWithDates.user_id
+WHERE DATEDIFF(order_date, prev_order_date) = 1
+AND user_id IN (
+    SELECT DISTINCT user_id
+    FROM UserOrdersWithDates
+    WHERE DATEDIFF(order_date, prev_order_date) = 1
+    AND user_id IN (
+        SELECT DISTINCT user_id
+        FROM UserOrdersWithDates
+        WHERE DATEDIFF(order_date, prev_order_date) = 1
+    )
+);
+