Aina Merchant - Data Engineering Assessment Shopify

sql/task1.sql

@@ -1,14 +1,32 @@
 -- Problem 1: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
 
+SELECT *
+FROM Products
+WHERE category_id = (SELECT category_id FROM Categories WHERE 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 Users.user_id, Users.username,
+  (SELECT COUNT(*) FROM Orders WHERE Orders.user_id = Users.user_id) AS total_orders
+FROM Users;
+
 -- 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 LEFT 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 LEFT 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;

sql/task2.sql

@@ -3,17 +3,69 @@
 -- 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.
 
+-- Common Table Expression that finds the average rating for each product and ranks them
+WITH AvgRatings AS (
+  SELECT
+    p.product_id,
+    p.product_name,
+    AVG(r.rating) AS average_rating,
+    RANK() OVER (ORDER BY AVG(r.rating) DESC) AS rating_rank
+  FROM
+    Products p
+  LEFT JOIN
+    Reviews r ON p.product_id = r.product_id
+  GROUP BY
+    p.product_id, p.product_name
+)
+
+SELECT product_id, product_name, average_rating
+FROM AvgRatings
+WHERE rating_rank = 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 DISTINCT category_id
+  FROM Categories
+  EXCEPT
+  SELECT DISTINCT p.category_id
+  FROM Products p
+  JOIN Order_Items oi ON p.product_id = oi.product_id
+  JOIN Orders o ON oi.order_id = o.order_id
+  WHERE u.user_id = o.user_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.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.
+
+-- Common Table Expression that finds the previous order date for each user
+WITH UserConsecutiveOrders 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 UserConsecutiveOrders uco ON u.user_id = uco.user_id
+WHERE DATEDIFF(day, uco.prev_order_date, uco.order_date) = 1
+ORDER BY u.user_id;

sql/task3.sql

@@ -3,17 +3,66 @@
 -- 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, 
+    COALESCE(SUM(oi.quantity * oi.unit_price), 0) AS total_sales_amount
+FROM Categories c
+LEFT JOIN Products p ON c.category_id = p.category_id
+LEFT JOIN Order_Items oi ON p.product_id = oi.product_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.
 
+-- Common Table Expression that gets the products belonging to the Toys & Games category
+WITH ToysAndGamesProducts AS (
+    SELECT product_id
+    FROM Products
+    WHERE category_id = (SELECT category_id FROM Categories WHERE category_name = 'Toys & Games')
+)
+SELECT u.user_id, u.username
+FROM Users u
+JOIN Orders o ON u.user_id = o.user_id
+JOIN Order_Items oi ON o.order_id = oi.order_id
+WHERE oi.product_id IN (SELECT product_id FROM ToysAndGamesProducts)
+GROUP BY u.user_id, u.username
+HAVING COUNT(DISTINCT oi.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.
 
+-- Common Table Expression that ranks on the basis of price for each product category
+WITH HighestPrice AS (
+    SELECT product_id, product_name, category_id, price, RANK() OVER (PARTITION BY category_id ORDER BY price DESC) AS rank
+    FROM Products
+)
+
+SELECT product_id, product_name, category_id, price
+FROM HighestPrice
+WHERE 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.
+
+-- Common Table Expression that calculates the start and end dates of three consecutive orders for each user 
+WITH UserOrderDates AS (
+  SELECT
+    user_id,
+    order_date AS start_date,
+    LEAD(order_date, 2) OVER (PARTITION BY user_id ORDER BY order_date) AS end_date
+  FROM Orders
+)
+
+SELECT DISTINCT
+  uod.user_id,
+  u.username
+FROM UserOrderDates uod
+JOIN Users u ON uod.user_id = u.user_id
+WHERE DATEDIFF(uod.end_date, uod.start_date) >= 2;