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Completed SQL FIles - Coding Assessment #185

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21 changes: 21 additions & 0 deletions sql/task1.sql
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@@ -1,14 +1,35 @@
-- Problem 1: Retrieve all products in the Sports category
-- Write an SQL query to retrieve all products in a specific category.
SELECT P.product_id, P.product_name, P.description, P.price, P.category_id
FROM Products P
JOIN Categories C ON P.category_id = C.category_id
WHERE C.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.
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
ORDER BY U.user_id;

-- 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
ORDER BY P.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_spent
FROM Users U
JOIN Orders O ON U.user_id = O.user_id
GROUP BY U.user_id, U.username
ORDER BY total_spent DESC
LIMIT 5;
62 changes: 61 additions & 1 deletion sql/task2.sql
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Expand Up @@ -3,17 +3,77 @@
-- 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.

DROP VIEW IF EXISTS ProductAverageRatings CASCADE;
CREATE VIEW ProductAverageRatings AS
SELECT product_id, AVG(rating) AS average_rating
FROM Reviews
GROUP BY product_id;

SELECT P.product_id, P.product_name, PAR.average_rating
FROM Products P
JOIN ProductAverageRatings PAR ON P.product_id = PAR.product_id
WHERE PAR.average_rating = (SELECT MAX(average_rating) FROM ProductAverageRatings);


-- 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.
DROP VIEW IF EXISTS UserCategories CASCADE;
DROP VIEW IF EXISTS UniqueCategories CASCADE;
CREATE VIEW UniqueCategories AS
SELECT DISTINCT category_id
FROM Categories;

CREATE VIEW UserCategories AS
SELECT DISTINCT U.user_id, P.category_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;

SELECT U.user_id, U.username
FROM Users U
WHERE NOT EXISTS (
SELECT 1
FROM UniqueCategories UC
WHERE NOT EXISTS (
SELECT 1
FROM UserCategories UC2
WHERE UC2.user_id = U.user_id AND UC2.category_id = UC.category_id
)
)
GROUP BY U.user_id, U.username;


-- 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.

DROP VIEW IF EXISTS UserOrderDates CASCADE;
DROP VIEW IF EXISTS ConsecutiveUserOrders CASCADE;

CREATE VIEW UserOrderDates AS
SELECT
user_id,
order_date,
LAG(order_date) OVER (PARTITION BY user_id ORDER BY order_date) AS previous_order_date
FROM Orders;
CREATE VIEW ConsecutiveUserOrders AS
SELECT
user_id
FROM UserOrderDates
WHERE order_date - previous_order_date = 1;
SELECT DISTINCT U.user_id, U.username
FROM Users U
JOIN ConsecutiveUserOrders CUO ON U.user_id = CUO.user_id;
85 changes: 85 additions & 0 deletions sql/task3.sql
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Expand Up @@ -3,17 +3,102 @@
-- 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(OI.quantity * OI.unit_price) AS total_sales_amount
FROM
Categories C
JOIN Products P ON C.category_id = P.category_id
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.
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(DISTINCT product_id)
FROM Products P
JOIN Categories C ON P.category_id = C.category_id
WHERE C.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.
DROP VIEW IF EXISTS HighestPricedProducts CASCADE;
CREATE VIEW HighestPricedProducts AS
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
FROM RankedProducts
WHERE price_rank = 1;
SELECT * FROM HighestPricedProducts;

-- 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.
DROP VIEW IF EXISTS UserConsecutiveOrders CASCADE;
DROP VIEW IF EXISTS UserConsecutiveStreaks CASCADE;
DROP VIEW IF EXISTS UserStreakLengths CASCADE;
DROP VIEW IF EXISTS UsersWith3DayStreaks CASCADE;
CREATE VIEW UserConsecutiveOrders AS
SELECT
user_id,
order_date,
LAG(order_date) OVER (PARTITION BY user_id ORDER BY order_date) AS previous_order_date
FROM Orders;
CREATE VIEW UserConsecutiveStreaks AS
SELECT
user_id,
order_date,
CASE
WHEN order_date - previous_order_date = 1 THEN 0
ELSE 1
END AS streak_start
FROM UserConsecutiveOrders;
CREATE VIEW UserStreakLengths AS
SELECT
user_id,
order_date,
SUM(streak_start) OVER (PARTITION BY user_id ORDER BY order_date) AS streak_id
FROM UserConsecutiveStreaks;
CREATE VIEW UsersWith3DayStreaks AS
SELECT
user_id,
COUNT(*) AS streak_length
FROM UserStreakLengths
GROUP BY user_id, streak_id
HAVING COUNT(*) >= 3;
SELECT DISTINCT U.user_id, U.username
FROM Users U
JOIN UsersWith3DayStreaks S ON U.user_id = S.user_id;
14 changes: 7 additions & 7 deletions tests/test_sql_queries.py
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Expand Up @@ -6,11 +6,11 @@ class TestSQLQueries(unittest.TestCase):
def setUp(self):
# Establish a connection to your test database
self.conn = psycopg2.connect(
dbname='your_dbname',
user='your_username',
password='your_password',
host='your_host',
port='your_port'
dbname='postgres',
user='postgres',
password='toor',
host='127.0.0.1',
port='5432'
)
self.cur = self.conn.cursor()

Expand All @@ -21,7 +21,7 @@ def tearDown(self):

def test_task1(self):
# Task 1: Example SQL query in task1.sql
with open('/sql/task1.sql', 'r') as file:
with open('./sql/task1.sql', 'r') as file:
sql_query = file.read()

self.cur.execute(sql_query)
Expand All @@ -36,7 +36,7 @@ def test_task1(self):

def test_task2(self):
# Task 2: Example SQL query in task2.sql
with open('/sql/task2.sql', 'r') as file:
with open('./sql/task2.sql', 'r') as file:
sql_query = file.read()

self.cur.execute(sql_query)
Expand Down