Carlos Subqueries

solutions.sql

@@ -1 +1,126 @@
--- Add you solution queries below:
+USE sakila;
+
+/* 1. How many copies of the film Hunchback Impossible exist in the inventory system? */
+SELECT film.title, COUNT(inventory.film_id)
+FROM film
+INNER JOIN inventory
+ON film.film_id = inventory.film_id
+WHERE title = "Hunchback Impossible"
+;
+
+
+/* 2. List all films whose length is longer than the average of all the films. */
+SELECT title, length
+FROM film
+WHERE length > 
+	(SELECT AVG(length) AS average 
+	FROM sakila.film)
+ORDER BY length DESC
+;
+
+
+/* 3. Use subqueries to display all actors who appear in the film Alone Trip. */
+SELECT DISTINCT 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 film.title = "Alone Trip"))
+;
+
+
+/* 4. Sales have been lagging among young families, and you wish to target all family movies for a promotion. Identify all movies categorized as family films. */
+SELECT film_id, title
+FROM film
+WHERE film_id in
+	(SELECT film_id
+    FROM film_category
+	WHERE category_id =
+		(SELECT category_id
+		FROM category
+		WHERE name = "Family"))
+;
+
+
+/* 5. Get name and email from customers from Canada using subqueries. Do the same with joins. Note that to create a join, 
+you will have to identify the correct tables with their primary keys and foreign keys, that will help you get the relevant information. */
+SELECT customer.first_name, customer.last_name, customer.email
+FROM customer
+WHERE customer.address_id IN
+	(SELECT address.address_id
+	FROM address
+	WHERE address.city_id IN
+		(SELECT city.city_id
+		FROM city
+		WHERE city.country_id =
+			(SELECT country.country_id
+			FROM country
+			WHERE country.country = "Canada")))
+;
+
+SELECT customer.first_name, customer.last_name, customer.email
+FROM customer
+INNER JOIN address
+ON customer.address_id = address.address_id
+INNER JOIN city
+ON address.city_id = city.city_id
+INNER JOIN country
+ON city.country_id = country.country_id
+WHERE country.country = "Canada"
+;
+
+
+/* 6. Which are films starred by the most prolific actor? Most prolific actor is defined as the actor that has acted in the most number of films. 
+First you will have to find the most prolific actor and then use that actor_id to find the different films that he/she starred. */
+SELECT film.title
+FROM film
+INNER JOIN film_actor
+ON film.film_id = film_actor.film_id
+WHERE film_actor.actor_id =
+	(SELECT actor.actor_id
+	FROM actor
+    INNER JOIN film_actor
+    ON actor.actor_id = film_actor.actor_id
+	GROUP BY film_actor.actor_id
+	ORDER BY COUNT(film_actor.film_id) DESC
+	LIMIT 1)
+;
+
+
+/* 7. Films rented by most profitable customer. You can use the customer table and payment table to find the most profitable customer 
+ie the customer that has made the largest sum of payments */
+SELECT DISTINCT film.title
+FROM film
+INNER JOIN inventory
+ON film.film_id = inventory.film_id
+INNER JOIN rental 
+ON inventory.inventory_id = rental.inventory_id
+INNER JOIN customer
+ON rental.customer_id = customer.customer_id
+WHERE customer.customer_id =
+	(SELECT customer.customer_id
+	FROM customer
+	INNER JOIN payment
+	ON customer.customer_id = payment.customer_id
+	GROUP BY customer.customer_id
+	ORDER BY SUM(payment.amount) DESC
+	LIMIT 1)
+;
+
+
+/* 8. Get 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. */
+SELECT DISTINCT customer.customer_id, SUM(payment.amount) AS total_amount_spent
+FROM customer
+INNER JOIN payment
+ON customer.customer_id = payment.customer_id
+GROUP BY customer.customer_id
+HAVING total_amount_spent >
+	(SELECT AVG(total_spent)
+	FROM
+		(SELECT SUM(payment.amount) AS total_spent
+		FROM payment
+		GROUP BY payment.customer_id) AS average_total)
+;