submission

sql/task1.sql

@@ -1,14 +1,36 @@
 -- Problem 1: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
+SELECT p.*, c.category_name
+FROM Products AS p LEFT JOIN Categories AS c ON p.category_id = c.category_id
+WHERE c.category_name = 'Sports & Outdoors';
 
 -- 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(distinct order_id) as total_num_of_orders
+FROM Users AS u LEFT JOIN Orders AS 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.
 
+# Exclude Review for Products Not in the Product Table as Products Table generally defines the set 
+# of valid produc
+SELECT p.product_id, p.product_name, AVG(rating) AS average_rating
+FROM Products AS p LEFT JOIN Reviews AS 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(total_amount) as total_amount_spent
+FROM Users AS u LEFT JOIN Orders as o ON u.user_id = o.order_id 
+GROUP BY u.user_id, u.username
+ORDER BY SUM(total_amount) DESC
+LIMIT 5;
+
+
+

sql/task2.sql

@@ -3,17 +3,54 @@
 -- 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(rating) as average_rating
+FROM Products as p LEFT JOIN Reviews as r ON p.product_id  = r.product_id 
+GROUP BY p.product_id, p.product_name 
+HAVING AVG(rating) = 
+#Find the highest average rating 
+(SELECT MAX(average_rating) 
+FROM (
+SELECT p.product_id, p.product_name, AVG(rating) as average_rating
+FROM Products as p LEFT JOIN Reviews as r ON p.product_id  = r.product_id 
+GROUP BY p.product_id, p.product_name ) AS p1) ;
+
 -- 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, COUNT(DISTINCT p.category_id) as category_count
+FROM Users as u LEFT JOIN Orders AS o on u.user_id = o.user_id
+	            LEFT JOIN Order_Items as i on o.order_id = i.order_id 
+	            LEFT JOIN Products as p on i.product_id = p.product_id
+GROUP BY u.user_id, u.username
+HAVING COUNT(DISTINCT p.category_id) = 
+(#retrive all the possible category_id from the Categories table
+SELECT COUNT(DISTINCT category_id)
+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, COUNT(review_id) AS num_review
+FROM Products AS p LEFT JOIN Reviews AS r on p.product_id = r.product_id
+GROUP BY p.product_id, p.product_name
+HAVING COUNT(review_id) = 0;
+
 
 -- 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.
+
+#Find the previous order date for each order 
+With ConsecutiveOrder AS (
+SELECT u.user_id, u.username, o.order_date, LAG(o.order_date) OVER (PARTITION BY u.user_id ORDER BY o.order_date) as previous_order_date
+FROM Users AS u LEFT JOIN Orders AS O ON u.user_id = o.user_id) 
+
+#If the date difference between the order date and the previous order date is equal to 1
+# we consider ths user have made orders in consecutive date
+SELECT DISTINCT user_id, username 
+FROM ConsecutiveOrder
+WHERE DATEDIFF(order_date, previous_order_date) = 1;

sql/task3.sql

@@ -3,17 +3,82 @@
 -- 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.
 
+SELECT c.category_id, c.category_name, SUM(quantity* unit_price) AS total_sales_amount
+FROM Categories AS c LEFT JOIN Products AS p ON c.category_id = p.category_id
+                     LEFT JOIN Order_Items AS i ON p.product_id = i.product_id
+GROUP BY c.category_id, c.category_name
+ORDER BY SUM(quantity* unit_price) 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.
 
+#Find the category_id for 'Toys & Games'
+WITH ToyCategory AS (
+SELECT category_id
+FROM Categories
+WHERE category_name = 'Toys & Games'), 
+
+#Find all product_id in that category
+ToyProductIDs AS (
+SELECT DISTINCT product_id
+FROM Products
+WHERE category_id = (SELECT * FROM ToyCategory)),
+
+#Find all orders
+AllOrders AS(
+SELECT u.user_id, u.username, i.product_id
+FROM Users as u JOIN Orders AS o ON u.user_id = o.user_id 
+     JOIN Order_Items AS i ON o.order_id = i.order_id 
+     JOIN Products AS p on p.product_id = i.product_id) 
+
+#Find all the users who have placed orders for all products in the Toys & Games
+SELECT user_id, username
+FROM ALLOrders
+GROUP BY user_id, username 
+HAVING COUNT(DISTINCT product_id) = (SELECT COUNT(*) FROM ToyProductIDs);
+
+
 -- 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.
+SELECT product_id, product_name, category_id, price
+FROM (
+#Rank product price in a dese order for each category
+SELECT product_id, product_name, category_id, price, 
+       RANK() OVER (PARTITION BY category_id ORDER BY price DESC) as price_rank
+FROM Products) AS p1 
+WHERE price_rank =1 ;
+
 
 -- 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.
+
+# Gather order date, keep unique order_date for each user_id
+WITH uniqueOrderDate as(
+SELECT user_id, order_date
+FROM Orders
+GROUP BY user_id, order_date),
+
+# For consecutive dates, the value of grp will be the same because for all consecutive dates
+consecutiveGroup as (
+SELECT *, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY order_date) AS row_num, 
+DATE_SUB(order_date, INTERVAL ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY order_date) DAY) as grp
+FROM uniqueOrderDate),
+
+# Select the user_id who have placed orders on consecutive days for at least 3 days
+userIdList AS (
+SELECT user_id, COUNT(*) AS consecutiveDates
+FROM consecutiveGroup
+GROUP BY user_id, grp
+HAVING COUNT(*) >=3) 
+
+#Get the distinct user_id and join Users table for user_name information
+SELECT DISTINCT l.user_id as user_id, u.username
+FROM userIdList AS l LEFT JOIN Users AS u ON l.user_id = u.user_id;
+