Completed all tasks

sql/task1.sql

@@ -1,14 +1,63 @@
 -- Problem 1: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
 
+-- CTE 'SportsCategory' retrieves category_ids for category names that contain 'sports' word 
+-- (case-insensitive incase sports is the second word)
+-- If 'Sportswear' was a category, this query would retrieve it's data as well
+
+WITH SportsCategory AS (
+    SELECT category_id
+    FROM Categories
+    WHERE LOWER(category_name) LIKE '%sports%'
+)
+
+SELECT *
+FROM Products
+WHERE category_id IN (SELECT category_id FROM SportsCategory);
+
+
 -- 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.
 
+-- Used a 'LEFT JOIN' to include all users (even those that may have place no orders)
+-- In PostgresSQL, COUNT() function doesn't return NULL if there are no rows to count.
+-- Instead, it returns zero.
+
+SELECT u.user_id, u.username, count(distinct o.order_id) as Total_Number_Orders
+FROM Users u
+LEFT JOIN Orders o ON o.user_id = u.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.
 
+
+-- Used a 'LEFT JOIN' to include all products (even those without any reviews)
+-- Grouping by p.product_id, p.product_nam, we consider all ratings for a specific product to calculate "average_rating"
+SELECT p.product_id, p.product_name, AVG(r.rating) AS average_rating
+FROM Products p 
+LEFT JOIN Reviews r ON r.product_id = p.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.
+
+-- For this query, I considered using the 'Payments' table but after observing that payment amount and order amount for user_id = 2 are different
+-- I'm assuming that Payment amount can include shipping costs, customs, etc.
+-- For this query, I am only considering the amount spent on the actual 'Order' (from Orders table)
+
+-- Here, I have used 'LEFT JOIN' to include all users (even those that may have place no orders)
+-- In PostgresSQL, The SUM() of an empty set will return NULL, not zero.
+-- Hence, I have used the COALESCE() function to return zero instead of NULL in case there is no matching row.
+-- The 'LIMIT 5' clause restricts the result to the top 5 users.
+SELECT u.user_id, u.username, COALESCE(SUM(o.total_amount), 0) AS total_amount_spent
+FROM Users u
+LEFT JOIN Orders o ON o.user_id = u.user_id
+GROUP BY u.user_id, u.username
+ORDER BY total_amount_spent DESC
+LIMIT 5;

sql/task2.sql

@@ -3,17 +3,59 @@
 -- 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.
 
+
+SELECT p.product_id, p.product_name, AVG(r.rating) AS average_rating
+FROM Products p
+LEFT JOIN Reviews r on r.product_id = p.product_id
+GROUP BY p.product_id, p.product_name
+ORDER BY 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.
 
+-- Counts the distinct categories for each user and compares it to the total count of categories in 'Categories'
+-- If equal, then user has made atleast one order in each category
+SELECT u.user_id, u.username
+FROM Users u
+JOIN Orders o ON o.user_id = u.user_id
+JOIN Order_Items oi on oi.order_id = o.order_id
+JOIN Products p on p.product_id = oi.product_id
+GROUP BY u.user_id, u.username
+HAVING COUNT(DISTINCT p.category_id) = (SELECT COUNT(*) FROM Categories);
+
+
+
+
 -- 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 r.product_id = p.product_id
+WHERE r.review_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.
+
+-- Calculates new column 'prev_order_date' for each order
+WITH PrevOrders AS (
+  SELECT user_id, order_id, order_date, LAG(order_date) OVER (PARTITION BY user_id ORDER BY order_date) AS prev_order_date
+  FROM Orders
+)
+-- Select only those users who have made consecutive orders on consecutive days
+SELECT DISTINCT u.user_id, u.username
+FROM Users u
+JOIN PrevOrders o ON u.user_id = o.user_id
+WHERE o.order_date = o.prev_order_date + INTERVAL 1 DAY;

sql/task3.sql

@@ -3,17 +3,81 @@
 -- 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.
 
+-- Calculates total sales amount for each category and then selects the top 3 categories
+SELECT C.category_id, C.category_name, SUM(O.total_amount) AS total_sales_amount
+FROM Categories C
+JOIN Products P ON C.category_id = P.category_id
+JOIN Order_Items OI ON P.product_id = OI.product_id
+JOIN Orders O ON OI.order_id = O.order_id
+GROUP BY C.category_id, C.category_name
+ORDER BY total_sales_amount 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.
 
+WITH ToysAndGamesProducts AS (
+  SELECT DISTINCT p.product_id
+  FROM Products p
+  JOIN Categories c ON p.category_id = c.category_id
+  WHERE c.category_name = 'Toys & Games'
+)
+SELECT u.user_id, u.username
+FROM Users u
+JOIN Orders o on o.user_id = u.user_id
+JOIN Order_Items oi on oi.order_id = o.order_id
+JOIN Products p on p.product_id = oi.product_id
+JOIN Categories c on c.category_id = p.category_id
+JOIN ToysAndGamesProducts tg ON p.product_id = tg.product_id
+GROUP BY u.user_id, u.username
+HAVING COUNT(DISTINCT p.product_id) =  (SELECT COUNT(*) FROM ToysAndGamesProducts);
+
+
+
+
+
 -- 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.
 
+
+
+-- This gives the max_price within each category
+WITH MaxPrice AS (
+    SELECT category_id, MAX(price) as max_price
+    FROM Products
+    GROUP BY category_id
+)
+
+SELECT p.product_id, p.product_name, mp.category_id, p.price
+FROM Products p 
+JOIN MaxPrice 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.
+
+-- Calculates two new column 'order_date2' and 'order_date3' for each order
+WITH ThreeDaysOrders AS (
+  SELECT
+    o.user_id, o.order_date,
+    LAG(o.order_date, 1) OVER (PARTITION BY o.user_id ORDER BY o.order_date) AS order_date2,
+    LAG(o.order_date, 2) OVER (PARTITION BY o.user_id ORDER BY o.order_date) AS order_date3
+  FROM Orders o
+)
+-- Only retrieve users for whom the current order date is one day after the second-order date and two days after the third-order date
+SELECT u.user_id, u.username
+FROM Users u
+JOIN ThreeDaysOrders o ON u.user_id = o.user_id
+WHERE o.order_date2 = DATE_SUB(o.order_date, INTERVAL 1 DAY)
+  AND o.order_date3 = DATE_SUB(o.order_date, INTERVAL 2 DAY);
+
+
+