From a3e20208e74fc54849273dd50f6bd1f6dd628e8e Mon Sep 17 00:00:00 2001
From: Caroline Berto Araujo <159922231+cbertopt@users.noreply.github.com>
Date: Sun, 6 Oct 2024 00:44:37 +0100
Subject: [PATCH] Update solutions.sql

---
 solutions.sql | 93 +++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 93 insertions(+)

diff --git a/solutions.sql b/solutions.sql
index d0eddcc..0d9127b 100644
--- a/solutions.sql
+++ b/solutions.sql
@@ -1 +1,94 @@
 -- Add you solution queries below:
+
+USE sakila;
+
+SELECT*
+FROM sakila.film;
+
+-- 1. How many copies of the film "Hunchback Impossible" exist in the inventory system?
+SELECT COUNT(*) AS total_copies
+FROM film f
+JOIN inventory 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 films.
+SELECT title, length
+FROM film
+WHERE length > (SELECT AVG(length) FROM film);
+
+-- 3. Use subqueries to display all actors who appear in the film "Alone Trip".
+SELECT first_name, last_name
+FROM actor
+WHERE actor_id IN (
+    SELECT actor_id
+    FROM film_actor
+    WHERE film_id = (SELECT film_id FROM film WHERE title = 'Alone Trip')
+);
+
+-- 4. Identify all movies categorized as family films.
+SELECT f.title
+FROM film f
+JOIN film_category fc ON f.film_id = fc.film_id
+JOIN category c ON fc.category_id = c.category_id
+WHERE c.name = 'Family';
+
+-- 5. Get name and email from customers from Canada using subqueries. Do the same with joins.
+SELECT first_name, last_name, email
+FROM customer
+WHERE address_id IN (
+    SELECT address_id
+    FROM address
+    WHERE city_id IN (
+        SELECT city_id
+        FROM city
+        WHERE country_id = (SELECT country_id FROM country WHERE country = 'Canada')
+    )
+);
+
+SELECT c.first_name, c.last_name, c.email
+FROM customer c
+JOIN address a ON c.address_id = a.address_id
+JOIN city ct ON a.city_id = ct.city_id
+JOIN country co ON ct.country_id = co.country_id
+WHERE co.country = 'Canada';
+
+-- 6. Which are films starred by the most prolific actor?
+WITH prolific_actor AS (
+    SELECT actor_id
+    FROM film_actor
+    GROUP BY actor_id
+    ORDER BY COUNT(film_id) DESC
+    LIMIT 1
+)
+SELECT f.title
+FROM film f
+JOIN film_actor fa ON f.film_id = fa.film_id
+WHERE fa.actor_id = (SELECT actor_id FROM prolific_actor);
+
+-- 7. Films rented by the most profitable customer.
+WITH top_customer AS (
+    SELECT customer_id
+    FROM payment
+    GROUP BY customer_id
+    ORDER BY SUM(amount) DESC
+    LIMIT 1
+)
+SELECT f.title
+FROM film f
+JOIN inventory i ON f.film_id = i.film_id
+JOIN rental r ON i.inventory_id = r.inventory_id
+WHERE r.customer_id = (SELECT customer_id FROM top_customer);
+
+-- 8. Get the client_id and the total_amount_spent of those clients who spent more than the average total amount spent by each client.
+WITH customer_spending AS (
+    SELECT customer_id, SUM(amount) AS total_spent
+    FROM payment
+    GROUP BY customer_id
+),
+average_spent AS (
+    SELECT AVG(total_spent) AS avg_spent
+    FROM customer_spending
+)
+SELECT customer_id, total_spent
+FROM customer_spending
+WHERE total_spent > (SELECT avg_spent FROM average_spent);
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