solutions

solutions.sql

@@ -1 +1,125 @@
 -- Add you solution queries below:
+-- 1. How many copies of the film 'Hunchback Impossible' exist in the inventory system?
+SELECT COUNT(*) AS copies_count
+FROM inventory
+WHERE film_id = (
+    SELECT film_id 
+    FROM film 
+    WHERE title = 'Hunchback Impossible'
+);
+
+-- 2. List all films whose length is longer than the average of all the films.
+SELECT film_id, title, length
+FROM film
+WHERE length > (
+    SELECT AVG(length)
+    FROM film
+)
+ORDER BY length DESC;
+
+-- 3. Use subqueries to display all actors who appear in the film 'Alone Trip'.
+SELECT actor_id, 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'
+    )
+)
+ORDER BY last_name, first_name;
+
+-- 4. Identify all movies categorized as family films.
+SELECT film_id, title, description
+FROM film
+WHERE film_id IN (
+    SELECT film_id
+    FROM film_category
+    WHERE category_id = (
+        SELECT category_id
+        FROM category
+        WHERE name = 'Family'
+    )
+)
+ORDER BY title;
+
+-- 5a. Get name and email from customers from Canada using subqueries.
+SELECT customer_id, 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'
+        )
+    )
+)
+ORDER BY last_name, first_name;
+
+-- 5b. Get name and email from customers from Canada using joins.
+SELECT c.customer_id, c.first_name, c.last_name, c.email
+FROM customer c
+JOIN address a ON c.address_id = a.address_id
+JOIN city ci ON a.city_id = ci.city_id
+JOIN country co ON ci.country_id = co.country_id
+WHERE co.country = 'Canada'
+ORDER BY c.last_name, c.first_name;
+
+-- 6. Which are films starred by the most prolific actor?
+WITH most_prolific_actor AS (
+    SELECT actor_id
+    FROM film_actor
+    GROUP BY actor_id
+    ORDER BY COUNT(film_id) DESC
+    LIMIT 1
+)
+SELECT f.film_id, f.title, f.description
+FROM film f
+WHERE f.film_id IN (
+    SELECT film_id
+    FROM film_actor
+    WHERE actor_id = (
+        SELECT actor_id
+        FROM film_actor
+        GROUP BY actor_id
+        ORDER BY COUNT(film_id) DESC
+        LIMIT 1
+    )
+)
+ORDER BY f.title;
+
+-- 7. Films rented by most profitable customer.
+SELECT DISTINCT f.film_id, 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 payment
+    GROUP BY customer_id
+    ORDER BY SUM(amount) DESC
+    LIMIT 1
+)
+ORDER BY f.title;
+
+-- 8. Get the client_id and the total_amount_spent of those clients who spent more than 
+SELECT customer_id AS client_id, 
+       SUM(amount) AS total_amount_spent
+FROM payment
+GROUP BY customer_id
+HAVING SUM(amount) > (
+    SELECT AVG(total_per_customer)
+    FROM (
+        SELECT customer_id, SUM(amount) AS total_per_customer
+        FROM payment
+        GROUP BY customer_id
+    ) AS customer_totals
+)
+ORDER BY total_amount_spent DESC;