Subqueries by Enrique

solutions.sql

@@ -1 +1,109 @@
 -- Add you solution queries below:
+use sakila;
+
+-- 1. How many copies of the film _Hunchback Impossible_ exist in the inventory system?
+select 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 title 
+FROM film 
+WHERE length > (SELECT AVG(length) FROM film);
+
+-- 3. Use subqueries to display all actors who appear in the film _Alone Trip_.
+
+select 
+    CONCAT(ac.first_name, ' ', ac.last_name) as Name_Actor,
+    f.title
+from film f 
+LEFT JOIN 
+    film_actor fa on f.film_id = fa.film_id
+inner JOIN
+    actor ac on fa.actor_id = ac.actor_id
+where f.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 
+    f.title,
+    ca.name 
+from film f
+left join 
+    film_category fc on f.film_id = fc.film_id
+inner join 
+    category ca on fc.category_id = ca.category_id
+where ca.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 
+    CONCAT(c.first_name, ' ', c.last_name, ' ', c.email) as Name_Customer,
+    co.country
+from customer c
+left join
+    address ad on c.address_id = ad.address_id
+inner join
+    city ci on ad.city_id = ci.city_id
+inner join
+    country co on ci.country_id = co.country_id
+where  co.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
+    f.title,
+    CONCAT(a.first_name, ' ', a.last_name) AS actor_name,
+    (SELECT COUNT(*) FROM film_actor WHERE actor_id = a.actor_id) AS total_films
+FROM film f
+INNER JOIN film_actor fa ON f.film_id = fa.film_id
+INNER JOIN actor a ON fa.actor_id = a.actor_id
+WHERE a.actor_id = (
+    SELECT actor_id
+    FROM film_actor
+    GROUP BY actor_id
+    ORDER BY COUNT(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 
+    concat(c.first_name, ' ', c.last_name) as customer_name,
+    sum(p.amount) as total_amount,
+    count(f.film_id) as films_rented
+from customer c
+inner join 
+    payment p on c.customer_id = p.customer_id
+inner join
+    rental r on p.rental_id = r.rental_id
+inner join 
+    inventory v on r.inventory_id = v.inventory_id
+inner join
+    film f on v.film_id = f.film_id
+group by concat(c.first_name, ' ', c.last_name)
+order by total_amount
+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.
+
+WITH customer_totals AS (
+    SELECT 
+        c.customer_id,
+        SUM(p.amount) AS total_amount_spent
+    FROM customer c
+    INNER JOIN payment p ON c.customer_id = p.customer_id
+    GROUP BY c.customer_id
+)
+SELECT 
+    customer_id,
+    total_amount_spent
+FROM customer_totals
+WHERE total_amount_spent > (SELECT AVG(total_amount_spent) FROM customer_totals)
+ORDER BY total_amount_spent DESC;