lab resuelto

lab_sql_subqueries.sql

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+-- Write SQL queries to perform the following tasks using the Sakila database:
+use sakila;
+-- 1 Determine the number of copies of the film "Hunchback Impossible" that exist in the inventory system.
+SELECT COUNT(si.inventory_id) as numero_de_copias
+FROM sakila.inventory as si
+WHERE si.film_id = (SELECT sf.film_id FROM sakila.film as sf WHERE sf.title ="Hunchback Impossible");
+
+-- 2 List all films whose length is longer than the average length of all the films in the Sakila database.
+SELECT sf.title
+FROM sakila.film as sf
+WHERE length > (SELECT AVG(length) FROM sakila.film);
+-- 3 Use a subquery to display all actors who appear in the film "Alone Trip".
+SELECT a.first_name,
+		a.last_name
+FROM sakila.actor as a
+WHERE a.actor_id IN (SELECT sfa.actor_id FROM sakila.film_actor as sfa 
+					WHERE sfa.film_id= (SELECT sf.film_id  FROM sakila.film as sf WHERE sf.title = "Alone Trip"));
+-- Bonus:
+-- 4 Sales have been lagging among young families, and you want to target family movies for a promotion. Identify all movies categorized as family films.
+
+
+SELECT sf.title
+FROM sakila.film as sf
+WHERE sf.film_id IN (SELECT sfc.film_id FROM sakila.film_category as sfc 
+					 WHERE sfc.category_id = (SELECT sc.category_id FROM sakila.category as sc WHERE name = "Family"));
+
+-- 5 Retrieve the name and email of customers from Canada using both subqueries and joins. To use joins, you will need to identify the relevant tables and their primary and foreign keys.
+SELECT scu.first_name, scu.last_name, scu.email
+FROM sakila.customer as scu
+WHERE address_id IN(SELECT sa.address_id FROM sakila.address as sa 
+					JOIN sakila.city as sc
+                    ON sa.city_id = sc.city_id
+                    JOIN sakila.country as cou
+                    ON sc.country_id=cou.country_id
+                    WHERE cou.country = "CANADA");
+
+-- 6 Determine which films were starred by the most prolific actor in the Sakila database. A prolific actor is defined as the actor who has acted in the most number of films. First, you will need to find the most prolific actor and then use that actor_id to find the different films that he or she starred in.
+
+SELECT title
+FROM sakila.film
+WHERE film_id IN (
+    SELECT film_id
+    FROM sakila.film_actor
+    WHERE actor_id = (
+        SELECT actor_id
+        FROM sakila.film_actor
+        GROUP BY actor_id
+        ORDER BY COUNT(film_id) DESC
+        LIMIT 1));
+
+-- 7 Find the films rented by the most profitable customer in the Sakila database. You can use the customer and payment tables to find the most profitable customer, i.e., the customer who has made the largest sum of payments.
+SELECT sf.title
+FROM sakila.film as sf
+JOIN sakila.inventory as si
+ON  sf.film_id=si.film_id
+WHERE si.film_id IN (SELECT si.film_id
+    FROM sakila.inventory as si
+    JOIN sakila.rental as sr
+    ON si.inventory_id = sr.inventory_id
+    WHERE sr.customer_id = (
+        SELECT sp.customer_id
+        FROM sakila.payment as sp
+        GROUP BY sp.customer_id
+        ORDER BY SUM(sp.amount) DESC
+        LIMIT 1))
+group by sf.title;
+-- 8 Retrieve 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. You can use subqueries to accomplish this.
+SELECT sp.customer_id, 
+       SUM(sp.amount) AS total_amount_spent
+FROM sakila.payment AS sp
+GROUP BY sp.customer_id
+HAVING SUM(sp.amount) > (
+    SELECT AVG(total_amount)
+    FROM (
+        SELECT SUM(amount) AS total_amount
+        FROM sakila.payment
+        GROUP BY customer_id
+    ) AS subquery
+);