Update solutions.sql

solutions.sql

@@ -1 +1,183 @@
 -- Add you solution queries below:
+-- 1. How many copies of the film _Hunchback Impossible_ exist in the inventory system?
+SELECT
+film.title,
+count(inventory.inventory_id) AS number_of_copies
+FROM film
+LEFT JOIN
+inventory on film.film_id = inventory.film_id
+WHERE film.title = "Hunchback Impossible"
+GROUP BY film.title;
+
+-- 2. List all films whose length is longer than the average of all the films.
+
+SELECT AVG(film.length)
+FROM film;
+
+SELECT
+film.title,
+film.length
+FROM film
+WHERE film.length > (
+SELECT AVG(film.length)
+FROM film)
+ORDER BY film.length ASC;
+
+
+-- 3. Use subqueries to display all actors who appear in the film _Alone Trip_.
+SELECT
+film.title,
+actor.actor_id,
+actor.first_name,
+actor.last_name
+FROM actor
+LEFT JOIN
+film_actor on actor.actor_id = film_actor.actor_id
+LEFT JOIN
+film on film_actor.film_id = film.film_id
+WHERE film.title = (
+SELECT film.title
+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,
+category.name
+from film
+left join 
+film_category on film.film_id = film_category.film_id
+left join
+category on film_category.category_id = category.category_id
+WHERE category.name = (
+SELECT category.name
+FROM category
+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.
+
+-- Subqueries
+
+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 IN (
+                SELECT country.country_id
+                FROM country
+                WHERE country = 'Canada'
+            )
+        )
+    );
+
+-- JOINS
+SELECT
+customer.first_name,
+customer.last_name,
+customer.email
+FROM customer
+LEFT JOIN
+address on customer.address_id = address.address_id
+LEFT JOIN
+city on address.city_id = city.city_id
+LEFT JOIN
+country on city.country_id = country.country_id
+WHERE 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 
+    a.first_name,
+    a.last_name,
+    f.title
+FROM 
+    film AS f
+JOIN 
+    film_actor AS fa ON f.film_id = fa.film_id
+JOIN 
+    actor AS a ON fa.actor_id = a.actor_id
+WHERE 
+    fa.actor_id = (
+        SELECT 
+            fa2.actor_id
+        FROM 
+            film_actor AS fa2
+        GROUP BY 
+            fa2.actor_id
+        ORDER BY 
+            COUNT(fa2.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 
+    f.title,
+    c.first_name,
+    c.last_name,
+    SUM(p.amount) AS total_payment
+FROM 
+    customer AS c
+JOIN 
+    payment AS p ON c.customer_id = p.customer_id
+JOIN 
+    rental AS r ON c.customer_id = r.customer_id
+JOIN 
+    inventory AS i ON r.inventory_id = i.inventory_id
+JOIN 
+    film AS f ON i.film_id = f.film_id
+WHERE 
+    c.customer_id = (
+        SELECT 
+            c2.customer_id
+        FROM 
+            customer AS c2
+        JOIN 
+            payment AS p2 ON c2.customer_id = p2.customer_id
+        GROUP BY 
+            c2.customer_id
+        ORDER BY 
+            SUM(p2.amount) DESC
+        LIMIT 1
+    )
+GROUP BY 
+    f.title, c.first_name, c.last_name
+ORDER BY 
+    total_payment DESC;
+
+-- 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 
+    c.customer_id AS client_id,
+    SUM(p.amount) AS total_amount_spent
+FROM 
+    customer AS c
+JOIN 
+    payment AS p ON c.customer_id = p.customer_id
+GROUP BY 
+    c.customer_id
+HAVING 
+    total_amount_spent > (
+        SELECT 
+            AVG(total_spent) 
+        FROM (
+            SELECT 
+                SUM(amount) AS total_spent
+            FROM 
+                payment
+            GROUP BY 
+                customer_id
+        ) AS subquery
+    );