My submission to Shopify intern assesment

import_script.py

@@ -0,0 +1,59 @@
+import pandas as pd
+import mysql.connector
+from mysql.connector import Error
+
+path = 'path/to/data'
+
+cart_data = 'cart_data.csv'
+cart_item_data = 'cart_item_data.csv'
+category_data = 'category_data.csv'
+order_data = 'order_data.csv'
+order_items_data = 'order_items_data.csv'
+payment_data = 'payment_data.csv'
+product_data = 'product_data.csv'
+review_data = 'review_data.csv'
+shipping_data = 'shipping_data.csv'
+user_data = 'user_data.csv'
+
+insert_order = [
+    (path + category_data, 'Categories'),
+    (path + product_data, 'Products'),
+    (path + user_data, 'Users'),
+    (path + order_data, 'Orders'),
+    (path + order_items_data, 'Order_Items'),
+    (path + review_data, 'Reviews'),
+    (path + cart_data, 'Cart'),
+    (path + cart_item_data, 'Cart_Items'),
+    (path + payment_data, 'Payments'),
+    (path + shipping_data, 'Shipping'),
+]
+
+
+def import_csv_to_mysql(csv_file, table_name, db_connection):
+    cursor = db_connection.cursor()
+
+    df = pd.read_csv(csv_file)
+
+    for i, row in df.iterrows():
+        sql = f"INSERT INTO {table_name} ({', '.join(df.columns)}) VALUES ({', '.join(['%s'] * len(df.columns))})"
+        try:
+            cursor.execute(sql, tuple(row))
+            db_connection.commit()
+        except Error as e:
+            print(f"Row {i} wasn't added: {e}")
+
+
+db_config = {
+    'host': 'localhost',
+    'database': 'db',
+    'user': 'root',
+    'password': 'password'
+}
+
+connection = mysql.connector.connect(**db_config)
+
+for data, table in insert_order:
+    import_csv_to_mysql(data, table, connection)
+
+if connection.is_connected():
+    connection.close()

sql/task1.sql

@@ -1,14 +1,50 @@
+-- TASK 1
+
 -- Problem 1: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
 
+-- COMMENT: Assumed there might be several sport categories, using pattern to match all of them
+
+SELECT *
+FROM Products
+WHERE category_id = (SELECT category_id 
+					 FROM Categories 
+                     WHERE category_name like '%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.
 
+-- COMMENT: left join because we need every user's data
+
+SELECT u.user_id, u.username, COUNT(o.order_id) AS total_orders
+FROM Users u
+LEFT 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.
 
+-- COMMENT: left join because we need every product's data
+
+SELECT p.product_id, p.product_name, AVG(r.rating) AS avg_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.
+
+-- COMMENT: used left join for the case when user with 0 orders makes it into top 5
+
+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

@@ -1,19 +1,66 @@
+-- TASK 2
+
 -- Problem 5: Retrieve the products with the highest average rating
 -- 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.
 
+-- COMMENT: comparing every product avg to max avg
+
+SELECT p.product_id, p.product_name, AVG(r.rating) AS avg_rating
+FROM Products p
+LEFT JOIN Reviews r ON p.product_id = r.product_id
+GROUP BY p.product_id, p.product_name
+HAVING avg_rating = (SELECT MAX(avg_rating1) 
+                     FROM (SELECT AVG(rating) AS avg_rating1 
+                           FROM Reviews 
+                           GROUP BY product_id) AS product_avg_ratings);
+
+
 -- 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.
 
+-- COMMENT: comparing user's categories count to total categories count
+
+SELECT Users.user_id, Users.username
+FROM Users
+WHERE Users.user_id IN (SELECT DISTINCT u.user_id 
+                        FROM Users u
+                        JOIN Orders o ON u.user_id = o.user_id
+                        JOIN Order_Items oi ON o.order_id = oi.order_id
+                        JOIN Products p ON oi.product_id = p.product_id
+                        GROUP BY u.user_id, p.category_id
+                        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 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.
+
+-- COMMENT: comparing consecutive rows order dates because looking for consecutive orders
+
+With ConsecutiveOrders AS (
+SELECT user_id,
+       order_date,
+       LEAD(order_date, 1) OVER (ORDER BY order_id) as next_order_date,
+       LEAD(user_id, 1) OVER (ORDER BY order_id) as next_user_id 
+FROM Orders
+)
+SELECT DISTINCT u.user_id, u.username
+FROM Users u
+JOIN ConsecutiveOrders co ON u.user_id = co.user_id
+WHERE co.user_id = co.next_user_id AND co.next_order_date = DATE_ADD(co.order_date, INTERVAL 1 DAY);

sql/task3.sql

@@ -1,19 +1,68 @@
+-- TASK 3
+
 -- Problem 9: Retrieve the top 3 categories with the highest total sales amount
 -- 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.
 
+-- COMMENT: here assumed unit price is an actual price for 1 item
+
+SELECT c.category_id, c.category_name, SUM(oi.unit_price * oi.quantity) 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.
 
+-- COMMENT: checking disctinct product_id count for each user for the Toys & Games category
+
+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
+JOIN Products p ON oi.product_id = p.product_id
+JOIN Categories c ON p.category_id = c.category_id
+WHERE c.category_name = 'Toys & Games'
+GROUP BY u.user_id, u.username
+HAVING COUNT(DISTINCT p.product_id) = (SELECT COUNT(*) 
+                                       FROM Products 
+                                       WHERE category_id = (SELECT category_id 
+                                                            FROM Categories 
+                                                            WHERE category_name = 'Toys & Games'));
+
+
 -- 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.
 
+-- COMMENT: taking row 1 from products ordered descending by price for each category
+
+SELECT product_id, product_name, category_id, price
+FROM (SELECT p.product_id, p.product_name, p.category_id, p.price, 
+      ROW_NUMBER() OVER (PARTITION BY p.category_id ORDER BY p.price DESC) AS row_num
+      FROM Products p) AS ranked_products
+WHERE row_num = 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.
+
+-- COMMENT: considering all combinations of 3 orders of the same user
+
+SELECT DISTINCT u.user_id, u.username
+FROM Users u
+JOIN Orders o1 ON u.user_id = o1.user_id
+JOIN Orders o2 ON u.user_id = o2.user_id
+JOIN Orders o3 ON u.user_id = o3.user_id
+WHERE o1.order_date = o2.order_date - INTERVAL 1 DAY 
+	AND o2.order_date = o3.order_date - INTERVAL 1 DAY;
+

test_data/cart_data.csv

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+cart_id,user_id
+1,1
+2,2
+3,3
+4,4
+5,5
+6,6
+7,7
+8,8
+9,9
+10,10
+11,11
+12,12
+13,13
+14,14
+15,15
+16,16
+17,17
+18,18
+19,19
+20,20
+21,21
+22,22
+23,23
+24,24
+25,25
+26,26
+27,27
+28,28
+29,29
+30,30

test_data/cart_item_data.csv

@@ -0,0 +1,37 @@
+cart_item_id,cart_id,product_id,quantity
+1,1,1,1
+2,1,2,2
+3,2,3,1
+4,2,4,3
+5,3,5,2
+6,3,6,1
+7,4,7,3
+8,4,8,1
+9,5,9,2
+10,5,10,1
+11,6,11,3
+12,6,12,1
+13,7,13,2
+14,7,14,1
+15,8,15,3
+16,8,16,1
+17,9,17,2
+18,9,18,1
+19,10,19,3
+20,10,20,1
+21,11,21,2
+22,11,22,1
+23,12,23,3
+24,12,24,1
+25,13,25,2
+26,13,26,1
+27,14,27,3
+28,14,28,1
+29,15,29,2
+30,15,30,1
+31,16,31,3
+32,16,32,1
+33,17,33,2
+34,17,34,1
+35,18,35,3
+36,18,36,1

test_data/category_data.csv

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+category_id,category_name
+1,Electronics
+2,Books
+3,Clothing
+4,Home & Kitchen
+5,Toys & Games
+6,Beauty & Personal Care
+7,Health & Household
+8,Sports & Outdoors
+9,Automotive
+10,Tools & Home Improvement
+11,Grocery & Gourmet Food
+12,Pet Supplies
+13,Office Products
+14,Music
+15,Movies & TV
+16,Arts
+17,Industrial & Scientific
+18,Electronics Accessories
+19,Cell Phones & Accessories
+20,Video Games
+21,Computers
+22,Appliances
+23,Software
+24,Kindle Store
+25,Home Audio & Theater
+26,Camera & Photo
+27,Shoes
+28,Jewelry
+29,Handmade
+30,CDs & Vinyl
+31,Baby
+32,Collectibles & Fine Art
+33,Instrument Accessories
+34,Power & Hand Tools
+35,Outdoor Recreation
+36,Home Décor
+37,Kitchen & Dining
+38,Health Care
+39,Office & School Supplies
+40,Industrial Electrical
+41,Industrial Hardware
+42,Industrial Power & Hand Tools
+43,Industrial Scientific
+44,Lab & Scientific Products
+45,Janitorial & Sanitation Supplies
+46,Test
+47,Occupational Health & Safety Products
+48,Science Education
+49,Raw Materials
+50,Food Service Equipment & Supplies