Lab subqueries

solutions.sql

@@ -1 +1,168 @@
--- Add you solution queries below:
+-- Solutions.sql -- 
+USE sakila;
+/*How many copies of the film Hunchback Impossible exist in the inventory system?*/
+SELECT 
+	film.title, COUNT(*) as copies
+    FROM film
+    INNER JOIN inventory
+    ON film.film_id = inventory.film_id
+WHERE film.title = "Hunchback Impossible"
+GROUP BY film.title;
+
+-- with subquery -- $$$ GIVES DIFFERENT NUMBER OF COPIES (4581) $$$ WWHYYYYYYY?
+Select 
+	COUNT(*) AS copies
+    FROM inventory
+    WHERE film_id = (
+    SELECT
+			film_id
+            FROM film_text
+		WHERE film_text.title = "Hunchback Impossible");
+
+;
+
+
+
+/*List all films whose length is longer than the average of all the films.*/
+SELECT
+	film.title, film.length
+    FROM sakila.film
+WHERE film.length > (
+    SELECT
+		AVG(film.length) AS Average_length 
+        FROM sakila.film)
+ORDER BY film.length;
+
+/*Use subqueries to display all actors who appear in the film Alone Trip.*/
+
+SELECT
+	actor.first_name, actor.last_name
+    FROM sakila.actor
+INNER JOIN (
+-- Subquery to find out the actor_id that matches the film_id from 'Alone Trip'
+    SELECT
+		film_actor.actor_id
+		FROM sakila.film_actor
+	INNER JOIN (
+    -- Subquery to get the film_id from title 'Alone Trip'
+		SELECT
+			film.film_id
+			FROM sakila.film
+		WHERE film.title = 'Alone Trip') AS selected_movie
+	ON sakila.film_actor.film_id = selected_movie.film_id) AS actors_in_selected_movie
+ON sakila.actor.actor_id = actors_in_selected_movie.actor_id
+ORDER BY actor.last_name asc;
+
+/*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 sakila.film
+INNER JOIN (
+    SELECT 
+		film_category.film_id
+		FROM sakila.film_category
+	INNER JOIN (
+		SELECT 
+			category.category_id
+			FROM sakila.category
+		WHERE category.name = 'Family') AS category_family
+	ON sakila.film_category.category_id = category_family.category_id) AS film_id_family
+ON sakila.film.film_id = film_id_family.film_id
+ORDER BY film.title;
+
+
+/*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 sakila.customer
+INNER JOIN (
+    SELECT
+		address.address_id
+		FROM sakila.address
+	INNER JOIN ( 
+    -- Subquery to get the city_id from country_id from "Canada")
+		SELECT
+			city.city_id
+			FROM sakila.city
+		INNER JOIN (
+        -- Subquery to get the country_id from "Canada"
+			SELECT
+				country.country_id
+				FROM sakila.country
+			WHERE country = 'Canada') AS country_selected
+		ON sakila.city.country_id = country_selected.country_id) AS city_id_canada -- country_id that matches in city
+        --
+	ON sakila.address.city_id = city_id_canada.city_id) AS address_id_canada -- city_id that matches in address
+ON sakila.customer.address_id = address_id_canada.address_id -- address_id that matches in customer
+;
+
+/*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
+	film.title
+    FROM sakila.film
+INNER JOIN (
+    -- subquery to find the film_id for all movies that match actor_id 
+	SELECT 
+		film_id, actor_id
+		FROM film_actor
+	WHERE film_actor.actor_id = (
+	-- Subquery to find the most prolifict actor 
+		SELECT DISTINCT	
+			film_actor.actor_id
+			FROM sakila.film_actor
+		GROUP BY actor_id
+		ORDER BY COUNT(film_actor.actor_id) desc
+		limit 1)
+	) AS films_top_actor
+ON sakila.film.film_id = films_top_actor.film_id
+;
+
+/*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 DISTINCT
+	film.title
+    FROM sakila.film 
+INNER JOIN (
+	-- Subquery to find the film ids from the films rented by the most profitable customer
+	SELECT 
+		film_id
+		FROM sakila.inventory
+	INNER JOIN (
+		-- Subquery to find the inventory ids of the films rented by the most profitable customer
+		SELECT 
+			rental.inventory_id
+			FROM sakila.rental
+		INNER JOIN (
+			-- Subquery to find the most profitable customer SUM(payment.amount)
+			SELECT 
+				payment.customer_id
+				from sakila.payment
+				GROUP BY customer_id
+				ORDER BY SUM(payment.amount) desc
+				limit 1) AS most_profitable_customer
+		ON rental.customer_id = most_profitable_customer.customer_id) AS inventory_selected
+	ON inventory.inventory_id = inventory_selected.inventory_id) AS films_selected
+ON film.film_id = films_selected.film_id
+ORDER BY film.title asc;
+
+
+/*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 distinct
+	customer_id, SUM(amount) AS total_amount_spent
+    FROM sakila.payment
+GROUP BY payment.customer_id
+HAVING SUM(amount) > (
+-- Subquery to find the Average spent by client
+	SELECT
+		SUM(amount) / COUNT(DISTINCT customer_id)
+		FROM sakila.payment) 
+ORDER BY SUM(amount) desc;
+