Kaiden Jessani - Shopify Application

README_FINAL

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+The "create_and_populate_database.py" program creates and populates a SQL database. 
+A results folder has been created. It has the expected results for each of the problems. 
+The "test_sql_queries.py" program executes the three SQL files, and compares the expected results against the actual results. The outcome is displayed in the console. 

results/problem1

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+product_id,product_name,description,price,category_id
+15,Mountain Bike,Conquer the trails with this high-performance mountain bike.,1000.0,8
+16,Tennis Racket,Take your tennis game to the next level with this professional-grade racket.,54.0,8

results/problem10

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+user_id,username
+5,sarahwilson

results/problem11

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+product_id,product_name,category_id,MAX(p.price)
+1,Smartphone X,1,500.0
+3,Laptop Pro,2,1200.0
+6,Designer Dress,3,300.0
+7,Coffee Maker,4,80.0
+9,Action Camera,5,200.0
+12,Skincare Set,6,150.0
+14,Weighted Blanket,7,100.0
+15,Mountain Bike,8,1000.0

results/problem12

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+user_id,username

results/problem2

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+user_id,username,count(o.order_id)
+1,johndoe,1
+2,janesmith,1
+3,maryjones,1
+4,robertbrown,1
+5,sarahwilson,1
+6,michaellee,1
+7,lisawilliams,1
+8,chrisharris,1
+9,emilythompson,1
+10,davidmartinez,1
+11,amandajohnson,1
+12,jasonrodriguez,1
+13,ashleytaylor,1
+14,matthewthomas,1
+15,sophiawalker,1
+16,jacobanderson,1
+17,olivialopez,1
+18,ethanmiller,1
+19,emilygonzalez,1
+20,williamhernandez,1
+21,sophiawright,1
+22,alexanderhill,1
+23,madisonmoore,1
+24,jamesrogers,1
+25,emilyward,1
+26,benjamincarter,1
+27,gracestewart,1
+28,danielturner,1
+29,elliecollins,1
+30,williamwood,1

results/problem3

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+product_id,product_name,AVG(r.rating)
+1,Smartphone X,5.0
+2,Wireless Headphones,4.0
+3,Laptop Pro,3.0
+4,Smart TV,5.0
+5,Running Shoes,2.0
+6,Designer Dress,4.0
+7,Coffee Maker,5.0
+8,Toaster Oven,3.0
+9,Action Camera,4.0
+10,Board Game Collection,1.0
+11,Yoga Mat,5.0
+12,Skincare Set,4.0
+13,Vitamin C Supplement,2.0
+14,Weighted Blanket,3.0
+15,Mountain Bike,5.0
+16,Tennis Racket,4.0

results/problem4

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+user_id,username,sum(o.total_amount)
+12,jasonrodriguez,160.0
+4,robertbrown,155.0
+8,chrisharris,150.0
+24,jamesrogers,150.0
+17,olivialopez,145.0

results/problem5

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+product_id,product_name,avg_rating
+1,Smartphone X,5.0
+4,Smart TV,5.0
+7,Coffee Maker,5.0
+11,Yoga Mat,5.0
+15,Mountain Bike,5.0

results/problem6

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+user_id,username

results/problem7

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+product_id,product_name

results/problem8

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+user_id,username

results/problem9

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+category_id,category_name,total_sales_amount
+8,Sports & Outdoors,155.0
+4,Home & Kitchen,145.0
+1,Electronics,125.0

sql/task1.sql

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--- Problem 1: Retrieve all products in the Sports category
+-- Problem 1:
+-- Question: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
+--
+-- Main Query:
+-- Selects all columns (p.*) from the 'Products' table.
+-- Uses a LEFT JOIN to include all products from the 'Products' table, and matching categories from the 'Categories' table based on the category_id.
+-- LEFT JOIN:
+-- Connects rows from 'Products' to 'Categories' based on the common column category_id.
+-- Products that do not have a matching category will still be included in the result with NULL values for category-related columns.
+-- WHERE clause:
+-- Filters the result to include only rows where the lowercase category_name from the 'Categories' table contains the substring 'sports'.
+-- The LIKE operator with % is used for a partial match, and LOWER() is used to perform a case-insensitive comparison.
+--
+-- Main query to select products from the 'Products' table based on a left join with 'Categories'
+SELECT p.* FROM Products p   -- Selects all columns from the 'Products' table
+    LEFT JOIN Categories c
+        ON p.category_id=c.category_id  ---- Left join with 'Categories' table
+    WHERE LOWER(c.category_name) like '%sports%'  -- Filters products where the category name contains 'sports' in a case-insensitive manner
 
--- Problem 2: Retrieve the total number of orders for each user
+-- *************************************************************************************************************************************************************
+-- Problem 2:
+-- Question: 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.
+--Main Query:
+-- Selects the user_id and username from the 'Users' table.
+-- Counts the number of orders for each user using the COUNT() function on the order_id.
+-- Uses a JOIN operation to connect rows from 'Users' to 'Orders' based on the common column user_id.
+-- JOIN clause:
+-- Connects rows from 'Users' to 'Orders' based on the common column user_id.
+-- GROUP BY clause:
+-- Groups the results by user_id, so the count of orders is calculated for each user separately.
+--
+-- Main query to count the number of orders for each user
+SELECT u.user_id, u.username, count(o.order_id) FROM Users u
+    JOIN Orders o
+        ON u.user_id=o.user_id  -- Joining Users and Orders tables based on user_id
+    GROUP BY u.user_id          -- Grouping the results by user_id
 
--- Problem 3: Retrieve the average rating for each product
+-- *************************************************************************************************************************************************************
+-- Problem 3:
+-- Question: 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.
+-- Main Query:
+-- Selects the product_id and product_name from the 'Products' table.
+-- Calculates the average rating for each product using the AVG() function on the rating column from the 'Reviews' table.
+-- Uses a JOIN operation to connect rows from 'Products' to 'Reviews' based on the common column product_id.
+-- JOIN clause:
+-- Connects rows from 'Products' to 'Reviews' based on the common column product_id.
+-- GROUP BY clause:
+-- Groups the results by product_id, so the average rating is calculated for each product separately.
+--
+-- Main query to calculate the average rating for each product based on reviews
+SELECT r.product_id, p.product_name, AVG(r.rating) FROM Products p
+    JOIN Reviews r
+        ON p.product_id=r.product_id  -- Joining Products and Reviews tables based on product_id
+    GROUP BY r.product_id             -- Grouping the results by product_id
 
+-- *************************************************************************************************************************************************************
 -- 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.
+--
+-- Main Query:
+-- Selects the user_id and username from the 'Users' table.
+-- Calculates the total amount spent by each user using the SUM() function on the total_amount column from the 'Orders' table.
+-- Uses a JOIN operation to connect rows from 'Users' to 'Orders' based on the common column user_id.
+-- JOIN clause:
+-- Connects rows from 'Users' to 'Orders' based on the common column user_id.
+--
+-- Groups the results by user_id and username, so the total amount is calculated for each user separately.
+-- Orders the results in descending order based on the total amount (3 represents the position of the SUM(o.total_amount) expression in the SELECT list).
+-- Limits the result set to the top 5 users by total amount spent.
+--
+-- Main query to calculate the total amount spent by each user and limit the result to the top 5 users by total amount
+SELECT u.user_id, u.username, sum(o.total_amount) FROM Users u
+    JOIN Orders o
+        ON u.user_id=o.user_id          -- Joining Users and Orders tables based on user_id
+    GROUP BY u.user_id, u.username      -- Grouping the results by user_id and username
+    ORDER BY 3 DESC                     -- Ordering the results in descending order based on total amount
+    LIMIT 5                             -- Limit the result to the top 5 users by total amount
+
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sql/task2.sql

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--- Problem 5: Retrieve the products with the highest average rating
+-- Problem 5:
+-- Question: 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.
+--
+-- avg_rating CTE Calculates the average ratings for each product using the AVG() function
+-- The PARTITION BY p.product_id ensures that the average is calculated for each product individually.
+WITH
+avg_rating AS (
+    -- Results to display product ID, product name, and the average rating
+    SELECT p.product_id, p.product_name, AVG(rating) over (PARTITION BY p.product_id) avg_rating FROM Products p
+        JOIN Reviews r
+            ON p.product_id=r.product_id
+)
+--Selects all columns from the AvgRating CTE.
+SELECT * FROM avg_rating
+    WHERE avg_rating = (SELECT MAX(avg_rating) FROM avg_rating) --include only rows where the avg_rating is equal to the maximum average rating
 
+-- *************************************************************************************************************************************************************
 -- 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.
+-- CTE DistinctCategories Calculates the count of distinct categories each user has ordered from.
+WITH
+DistinctCategories AS (
+    SELECT COUNT(DISTINCT c.category_id) dist_cat_n, u.user_id, u.username FROM Users u
+        JOIN Orders o  --Joins the Users, Orders, Order_Items, Products, and Categories tables
+            ON u.user_id=o.user_id
+        JOIN Order_Items oi
+            ON o.order_id=oi.order_id
+        JOIN Products p
+            ON p.product_id=oi.product_id
+        JOIN Categories c
+            ON p.category_id=c.category_id
+        GROUP BY 2,3     -- group the results by user_id and username
+)
+SELECT user_id, username FROM DistinctCategories
+WHERE dist_cat_n = (SELECT COUNT(DISTINCT category_id) FROM Categories) --filters the results to include only users who have ordered from all distinct categories
 
--- Problem 7: Retrieve the products that have not received any reviews
+-- *************************************************************************************************************************************************************
+-- Problem 7:
+-- Question: 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.
+--
+-- The Main Query Selects product_id and product_name from the Products table.
+-- The WHERE clause filters the results to include only products whose product_id is not present in the subquery.
+--
+-- Main query to select products that have no reviews
+SELECT product_id, product_name FROM Products
+WHERE product_id not in (SELECT DISTINCT product_id FROM Reviews) -- Subquery to get distinct product IDs from the Reviews table
 
--- Problem 8: Retrieve the users who have made consecutive orders on consecutive days
+-- *************************************************************************************************************************************************************
+-- Problem 8:
+-- Question: 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.
+--
+-- CTE orders_with_prev:
+-- Calculates the previous order date for each order of each user.
+-- The LAG window function is used to obtain the previous order date based on the order date for each user.
+-- Main Query:
+-- Selects user_id and username from the orders_with_prev CTE.
+-- The WHERE clause filters the results to include only users whose orders are on consecutive days,
+-- The CAST function is used to convert the Julian day differences to integers.
+--
+-- CTE named 'orders_with_prev' to calculate the previous order date for each order
+WITH orders_with_prev AS (
+    SELECT 
+        o.user_id, 
+        u.username, 
+        o.order_date, 
+        LAG(o.order_date,1) OVER (PARTITION BY o.user_id ORDER BY o.order_date) previous_order_date 
+    FROM Orders o
+        JOIN Users u
+            ON o.user_id=u.user_id
+        ORDER BY 1,3  -- Order by user_id and order_date
+    )
+-- Main query to select user_id and username from the 'orders_with_prev' CTE
+    SELECT 
+        user_id, 
+        username
+    FROM orders_with_prev
+    WHERE Cast((JulianDay(order_date) - JulianDay(previous_order_date)) As Integer) = 1 -- Filter users with consecutive orders