Subquery Lab MySQL

solutions.sql

@@ -1 +1,155 @@
 -- Add you solution queries below:
+-- 1. How many copies of the film _Hunchback Impossible_ exist in the inventory system?
+SELECT 
+    COUNT(*) AS "Number of Copies"
+FROM 
+    inventory 
+INNER JOIN 
+    film  ON inventory.film_id = film.film_id
+WHERE 
+    film.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) FROM film);
+
+-- 3. Use subqueries to display all actors who appear in the film _Alone Trip_.
+SELECT 
+    actor.first_name, 
+    actor.last_name
+FROM 
+    actor 
+WHERE 
+    actor.actor_id IN (
+        SELECT film_actor.actor_id
+        FROM film_actor 
+        WHERE film_actor.film_id = (
+            SELECT film.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.title
+FROM 
+    film
+INNER JOIN 
+    film_category ON film.film_id = film_category.film_id
+INNER JOIN 
+    category ON film_category.category_id = category.category_id
+WHERE 
+    category.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
+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.
+-- Most prolific actor
+SELECT 
+    actor_id,
+    COUNT(film_id) AS film_count
+FROM 
+    film_actor
+GROUP BY 
+    actor_id
+ORDER BY 
+    film_count DESC
+LIMIT 1;
+
+
+-- Films he's been on
+SELECT 
+    film.title
+FROM 
+    film
+INNER JOIN 
+    film_actor ON film.film_id = film_actor.film_id
+WHERE 
+    film_actor.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
+-- Most profitable customer
+SELECT 
+    customer_id,
+    SUM(amount) AS total_payment
+FROM 
+    payment
+GROUP BY 
+    customer_id
+ORDER BY 
+    total_payment DESC
+LIMIT 1;
+
+-- Films rented
+SELECT 
+    film.title
+FROM 
+    rental
+INNER JOIN 
+    inventory ON rental.inventory_id = inventory.inventory_id
+INNER JOIN 
+    film ON inventory.film_id = film.film_id
+WHERE 
+    rental.customer_id = (
+        SELECT 
+            customer_id
+        FROM 
+            payment
+        GROUP BY 
+            customer_id
+        ORDER BY 
+            SUM(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 
+    customer_id, 
+    SUM(amount) AS total_amount_spent
+FROM 
+    payment
+GROUP BY 
+    customer_id
+HAVING 
+    total_amount_spent > (
+        SELECT AVG(total_amount) 
+        FROM (
+            SELECT SUM(amount) AS total_amount 
+            FROM payment 
+            GROUP BY customer_id
+        ) AS avg_spent
+    );