Solved Lab

lab4.sql

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+USE sakila;
+-- Write SQL queries to perform the following tasks using the Sakila database:
+-- 1. Determine the number of copies of the film "Hunchback Impossible" that exist in the inventory system.
+SELECT COUNT(i.inventory_id) AS number_of_copies FROM sakila.film AS f
+JOIN sakila.inventory AS i
+ON f.film_id = i.film_id
+WHERE f.title = 'Hunchback Impossible';
+
+-- 2. List all films whose length is longer than the average length of all the films in the Sakila database.
+SELECT title, length FROM sakila.film
+WHERE length > (SELECT AVG(length) FROM sakila.film);
+
+-- 3. Use a subquery to display all actors who appear in the film "Alone Trip".
+SELECT a.first_name, a.last_name FROM sakila.actor AS a
+JOIN sakila.film_actor AS fa 
+ON a.actor_id = fa.actor_id
+WHERE fa.film_id = (SELECT f.film_id FROM sakila.film AS f WHERE f.title = 'Alone Trip');
+
+-- BONUS
+-- 4. Sales have been lagging among young families, and you want to target family movies for a promotion. Identify all movies categorized as family films.
+SELECT f.title FROM sakila.film AS f
+JOIN sakila.film_category AS fc 
+ON f.film_id = fc.film_id
+JOIN sakila.category AS c 
+ON fc.category_id = c.category_id
+WHERE c.name = 'Family';
+
+-- 5 Retrieve the name and email of customers from Canada using both subqueries and joins.
+-- To use joins, you will need to identify the relevant tables and their primary and foreign keys.
+
+-- using joins:
+SELECT cu.first_name, cu.last_name, cu.email FROM sakila.customer AS cu
+JOIN sakila.address AS a 
+ON cu.address_id = a.address_id
+JOIN sakila.city AS ci 
+ON a.city_id = ci.city_id
+JOIN sakila.country AS co 
+ON ci.country_id = co.country_id
+WHERE co.country = 'Canada';
+
+-- using subqueries:
+SELECT cu.first_name, cu.last_name, cu.email FROM sakila.customer AS cu
+WHERE cu.address_id IN (SELECT a.address_id FROM sakila.address AS a
+WHERE a.city_id IN (SELECT ci.city_id FROM sakila.city AS ci
+WHERE ci.country_id = (SELECT co.country_id FROM sakila.country AS co
+WHERE co.country = 'Canada')));
+
+-- 6. Determine which films were starred by the most prolific actor in the Sakila database. A prolific actor is defined as the actor who has acted in the most number of films.
+-- First, you will need to find the most prolific actor and then use that actor_id to find the different films that he or she starred in.
+SELECT actor_id FROM sakila.film_actor
+GROUP BY actor_id
+ORDER BY COUNT(film_id) DESC
+LIMIT 1; -- this is the most prolific actor
+
+-- and now lets find the films he starred
+SELECT f.title FROM sakila.film AS f
+JOIN sakila.film_actor AS fa 
+ON f.film_id = fa.film_id
+WHERE fa.actor_id = (SELECT actor_id FROM sakila.film_actor -- adding the most prolific actor
+GROUP BY actor_id
+ORDER BY COUNT(film_id) DESC
+LIMIT 1);
+
+-- 7. Find the films rented by the most profitable customer in the Sakila database. 
+-- You can use the customer and payment tables to find the most profitable customer, i.e., the customer who has made the largest sum of payments.
+SELECT customer_id FROM sakila.payment
+GROUP BY customer_id
+ORDER BY SUM(amount) DESC
+LIMIT 1; -- this is the most profitable customer
+
+-- now lets find the films he rented:
+SELECT DISTINCT f.title FROM sakila.film AS f
+JOIN sakila.inventory AS i 
+ON f.film_id = i.film_id
+JOIN sakila.rental AS r 
+ON i.inventory_id = r.inventory_id
+WHERE r.customer_id = (SELECT customer_id -- adding the most profitable customer
+FROM sakila.payment
+GROUP BY customer_id
+ ORDER BY SUM(amount) DESC
+LIMIT 1);
+
+-- 8. Retrieve the client_id and the total_amount_spent of those clients who spent more than the average of the total_amount spent by each client. 
+-- You can use subqueries to accomplish this.
+SELECT customer_id, SUM(amount) AS total_amount_spent FROM sakila.payment
+GROUP BY customer_id
+HAVING total_amount_spent > (SELECT AVG(total_spent) FROM (SELECT SUM(amount) AS total_spent FROM sakila.payment GROUP BY customer_id) AS subquery);