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Bilal data eng #209

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task 1 completed
Minato0859 Jan 30, 2024
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commetned task 1
Minato0859 Jan 30, 2024
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task 2, 3, 2 orginised, 3 rushed
Minato0859 Jan 30, 2024
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Minato0859 Jan 30, 2024
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24 changes: 24 additions & 0 deletions sql/task1.sql
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-- Problem 1: Retrieve all products in the Sports category
-- Write an SQL query to retrieve all products in a specific category.

SELECT * --selecting all columns
FROM Products --from the Products table
WHERE category_id = 1; --where the category_id is 1 (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.

SELECT u.user_id, u.username, COUNT(o.order_id) AS total_orders --selecting the user_id, username, and total number of orders
FROM Users AS u -- from the Users table
LEFT JOIN Orders AS o ON u.user_id = o.user_id --left joining the Orders table on the user_id
GROUP BY u.user_id, u.username; --grouping by the user_id and 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.

SELECT p.product_id, p.product_name, AVG(r.rating) AS average_rating --selecting the product_id, product_name, and average rating
FROM Products AS p --from the Products table
LEFT JOIN Reviews AS r ON p.product_id = r.product_id --left joining the Reviews table on the product_id
ORDER BY p.product_id; --ordering by the 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.

SELECT u.user_id, u.username, SUM(o.total_amount) AS total_amount_spent --selecting the user_id, username, and total amount spent
FROM Users AS u --from the Users table
JOIN Orders AS o ON u.user_id = o.user_id -- joining the Orders table on the user_id
GROUP BY u.user_id, u.username -- grouping by the user_id and username
ORDER BY total_amount_spent DESC -- ordering by the total amount spent in descending order
LIMIT 5; -- limiting the results to 5
78 changes: 77 additions & 1 deletion sql/task2.sql
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-- 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.

-- Create a CTE to calculate the average rating for each product
-- Create a CTE to rank the products by average rating
WITH ProductRatings AS (
SELECT
product_id,
AVG(rating) AS average_rating
FROM Reviews
GROUP BY product_id
)


, RankedProducts AS (

SELECT
p.product_id,
p.product_name,
pr.average_rating,
RANK() OVER (ORDER BY pr.average_rating DESC) AS rating_rank
FROM Products AS p
JOIN ProductRatings AS pr ON p.product_id = pr.product_id
)


SELECT -- Retrieve the products with the highest average rating
product_id,
product_name,
average_rating
FROM RankedProducts
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 users who have ordered from all categories
SELECT
u.user_id,
u.username
FROM
Users AS u
WHERE
-- Count the distinct categories from which the user has ordered
(SELECT COUNT(DISTINCT p.category_id)
FROM Orders AS o
JOIN Order_Items AS oi ON o.order_id = oi.order_id
JOIN Products AS p ON oi.product_id = p.product_id
WHERE o.user_id = u.user_id) =
-- Compare to the total number of categories available
(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 products that have not received any reviews
SELECT p.product_id, p.product_name
FROM Products AS p
LEFT JOIN Reviews AS r ON p.product_id = r.product_id
WHERE r.product_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.

-- Create a CTE to retrieve the users who have made orders on consecutive days
WITH OrderedUsers AS (
SELECT
u.user_id,
u.username,
o.order_date,
LAG(o.order_date) OVER (PARTITION BY o.user_id ORDER BY o.order_date) AS previous_order_date
FROM
Orders AS o
JOIN Users AS u ON o.user_id = u.user_id
)

SELECT DISTINCT -- Select users who have made consecutive orders on consecutive days
user_id,
username
FROM
OrderedUsers
WHERE
order_date = DATE(previous_order_date, '+1 day');


51 changes: 51 additions & 0 deletions sql/task3.sql
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-- 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 category details and calculate total sales amount
SELECT c.category_id, c.category_name, SUM(oi.quantity * oi.unit_price) AS total_sales_amount
FROM Categories c
JOIN Products p ON c.category_id = p.category_id -- Link products to their categories
JOIN Order_Items oi ON p.product_id = oi.product_id -- Get sales data for products
GROUP BY c.category_id, c.category_name -- Aggregate sales by category
ORDER BY total_sales_amount DESC -- Order categories by total sales
LIMIT 3; -- Return only the top 3 categories


-- 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 users who ordered all 'Toys & Games' products
SELECT u.user_id, u.username
FROM Users u
JOIN Orders o ON u.user_id = o.user_id -- Join orders to users
JOIN Order_Items oi ON o.order_id = oi.order_id -- Get items in those orders
JOIN Products p ON oi.product_id = p.product_id -- Link those items to products
WHERE p.category_id = (
SELECT category_id FROM Categories WHERE category_name = 'Toys & Games' -- Target 'Toys & Games' category
)
GROUP BY u.user_id, u.username -- Group by user, Ensure user ordered all products in category
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.

-- Rank products by price within each category
WITH RankedProducts AS (
SELECT product_id, product_name, category_id, price,
RANK() OVER (PARTITION BY category_id ORDER BY price DESC) AS price_rank
FROM Products
)


SELECT product_id, product_name, category_id, price -- Select highest-priced products in each category
FROM RankedProducts
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.
-- ee

-- Identify consecutive order days for each user
WITH consecutivedays AS (
SELECT user_id, order_date, DATE(order_date) - ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY order_date) AS grp FROM Orders
),
-- Count the length of each consecutive day streak
consec AS (
SELECT user_id, COUNT(*) AS streak_length FROM consecutivedays GROUP BY user_id, grp HAVING COUNT(*) >= 3
)
SELECT DISTINCT o.user_id, u.username FROM consec o JOIN Users u ON o.user_id = u.user_id;