W3D2 lab 3

solutions.sql

@@ -1 +1,81 @@
 -- Add you solution queries below:
+/* How many copies of the film Hunchback Impossible exist in the inventory system? */
+SELECT COUNT(inventory_id) AS total_copies
+FROM inventory
+WHERE film_id = (SELECT film_id FROM film WHERE title = 'Hunchback Impossible');
+
+/* List all films whose length is longer than the average of all the films. */
+SELECT title, length
+FROM film
+WHERE length > (SELECT AVG(length) FROM film);
+
+/* 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')
+);
+
+/* 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 title
+FROM film
+WHERE film_id IN (
+	SELECT film_id
+    FROM film_category
+    WHERE category_id = (SELECT category_id FROM category WHERE name = 'Family')
+);
+
+/* 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 c.first_name, c.last_name, c.email
+FROM customer c
+JOIN address a ON c.address_id = a.address_id
+JOIN city cy ON a.city_id = cy.city_id
+JOIN country coun ON cy.country_id = coun.country_id 
+WHERE coun.country = 'Canada';
+
+/* 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 actor_id, COUNT(film_id) AS  film_count
+FROM film_actor
+GROUP BY actor_id
+ORDER BY film_count DESC
+LIMIT 1;
+
+SELECT title
+FROM film
+WHERE film_id IN (
+	SELECT film_id
+    FROM film_actor
+	WHERE actor_id = 107
+);
+
+/* 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 customer_id
+FROM payment
+GROUP BY customer_id
+ORDER BY SUM(amount) DESC
+LIMIT 1;
+
+SELECT title
+FROM film
+WHERE film_id IN (
+	SELECT film_id
+    FROM rental 	
+    WHERE inventory_id IN (
+		SELECT inventory_id
+        FROM inventory
+        WHERE store_id IN (
+			SELECT store_id
+            FROM customer
+            WHERE customer_id = 526
+		)
+	)
+);
+
+/* 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 customer_id, SUM(amount) AS total_amount_spent
+FROM payment
+GROUP BY customer_id
+HAVING total_amount_spent > (SELECT AVG(total_amount_spent) FROM (SELECT SUM(amount) AS total_amount_spent FROM payment GROUP BY customer_id) AS avg_spending);