From 5c4f07188531caba8adc346b703a1f9777913acc Mon Sep 17 00:00:00 2001
From: Massih Chopan <194472108+FulSt-Mas@users.noreply.github.com>
Date: Fri, 11 Jul 2025 16:38:48 +0200
Subject: [PATCH] solutio

---
 solutions.sql | 146 ++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 146 insertions(+)

diff --git a/solutions.sql b/solutions.sql
index d0eddcc..31ed06d 100644
--- a/solutions.sql
+++ b/solutions.sql
@@ -1 +1,147 @@
 -- Add you solution queries below:
+
+use sakila;
+
+--  1. How many copies of the film Hunchback Impossible exist in the inventory system?
+-- Select coun.  , from Inveotry, where Film tiel i hunback Via film ID in film table 
+
+
+select * from inventory;
+select * from film;
+select * from store;
+
+
+Select 
+	f.title,
+    COUNT(i.inventory_id) as number_copies
+from film f
+JOIN inventory i on f.film_id = i.film_id
+where f.title = 'Hunchback Impossible'
+group by i.film_id;
+
+
+--  2. List all films whose length is longer than the average of all the films.
+
+SELECT 
+    f.film_id,
+    f.title,
+    f.length
+FROM film f
+WHERE f.length > (
+    SELECT AVG(length) FROM film
+);
+
+
+-- 3 Use subqueries to display all actors who appear in the film Alone Trip.
+
+-- Selce Acto id, conncat actor name  in in Film_actor, fitlm titel
+-- where about film tiel alonte w
+-- join film actor and film via film id and film actor id with film actor and acotr tbales 
+--
+
+Select 
+	f.title,
+    a.actor_id,
+    concat(a.first_name, ' ', a.last_name) as actor_name
+from film f
+Join film_actor fa on fa.film_id = f.film_id
+Join actor a on fa.actor_id = a.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,
+    c.name
+from film f
+join film_category fc on f.film_id =fc.film_id
+join category c on fc.category_id = c.category_id
+where c.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) as name,
+    c.email,
+    co.country
+from customer c
+join address a on c.address_id = a.address_id
+join city ci on a.city_id = ci.city_id 
+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.
+
+WITH TopActors AS (
+	Select 
+		a.actor_id,
+		concat(a.first_name, ' ', a.last_name) as name_prolific_actor,
+		count(fa.film_id) as total_films
+	From actor a 
+	join film_actor fa on a.actor_id = fa.actor_id
+	GROUP BY a.actor_id  -- Grouping by actor_id ensures each actor is counted separately
+	order by total_films desc
+    limit 10
+)
+SELECT
+    ta.actor_id AS id,
+    ta.name_prolific_actor,
+    ta.total_films,
+    GROUP_CONCAT(f.title) AS films
+FROM TopActors ta
+JOIN film_actor fa ON ta.actor_id = fa.actor_id
+JOIN film f ON fa.film_id = f.film_id
+GROUP BY ta.actor_id, ta.name_prolific_actor, ta.total_films
+order by total_films desc;
+
+--  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
+
+With profitable_customer AS (
+	select 
+		c.customer_id,
+		concat(c.first_name, ' ', c.last_name) as name,
+		sum(p.amount) as total_amount
+	from customer c 
+	Join payment p on c.customer_id = p.customer_id
+	group by c.customer_id
+	order by total_amount desc
+	limit 1
+)
+Select 
+	pc.customer_id as id,
+    pc.name,
+	GROUP_CONCAT(f.title) AS films
+from profitable_customer pc
+JOIN payment p ON pc.customer_id = p.customer_id  
+join rental r on p.rental_id = r.rental_id
+join inventory i on r.inventory_id = i.inventory_id
+join film f on i.film_id = f.film_id
+group by pc.customer_id, pc.name;
+
+
+--  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
+        p.customer_id,
+        SUM(p.amount) AS total_amount_spent
+    FROM payment p
+    GROUP BY p.customer_id
+)
+SELECT 
+    ct.customer_id,
+    ct.total_amount_spent
+FROM customer_totals ct
+WHERE ct.total_amount_spent > (
+    SELECT AVG(total_amount_spent) FROM customer_totals
+);
+
+	
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