week4 lab3.2 done

lab_3.2.sql

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+-- 1
+select count(*) as number_of_copie 
+from film as f
+join inventory as inv
+on f.film_id = inv.film_id
+where title = 'HUNCHBACK IMPOSSIBLE';
+
+-- 2
+select title, length
+from film
+where length > (select avg(length) from film);
+
+SELECT f.title, f.length
+FROM film AS f
+JOIN (
+    SELECT AVG(length) AS average_length
+    FROM film
+) AS avg_table
+ON f.length > avg_table.average_length;
+
+-- 3
+select concat(first_name, ' ', last_name) as ALONE_TRIP_full_name_actor
+from actor as a
+where a.actor_id in (
+	select fa.actor_id 
+    from film_actor as fa
+    where fa.film_id in (
+		select f.film_id
+        from film as f
+        where f.title = 'ALONE TRIP'
+        )
+);
+
+-- 4  identy all_families_mvies
+select f.title, c.name
+from film as f
+join film_category as fc on f.film_id = fc.film_id
+join category as c on fc.category_id = c.category_id
+where c.name in ('Children', 'Comedy', 'Family');
+
+-- 5 name_email_customer_canada
+select c.first_name, c.email 
+from customer as c
+join address as a on c.address_id = a.address_id
+join city as ci on a.city_id = ci.city_id
+where ci.country_id in (
+			select co.country_id
+            from country as co
+            where co.country = 'Canada');
+
+-- 6 film of the most prolific actor
+select f.title
+from film as f
+join film_actor as fa on f.film_id = fa.film_id
+where fa.actor_id = (
+	select fa.actor_id
+	from film_actor as fa
+	group by fa.actor_id
+	order by count(fa.actor_id) desc
+    limit 1);
+
+
+-- 7 customer + de payment - les films loués par lui 
+
+select distinct(f.title)
+from rental as r
+join payment as p on r.customer_id = p.customer_id
+join inventory as inv on r.inventory_id = inv.inventory_id
+join film  as f on inv.film_id = f.film_id
+where r.customer_id = (
+	select p.customer_id
+    from payment as p
+    group by p.customer_id
+    order by sum(p.amount) desc
+    limit 1);
+
+
+-- 8 client_id/amount_spent pour client > avg spent
+
+select c.customer_id, sum(p.amount) as total_spent
+from customer c
+join payment p on c.customer_id = p.customer_id
+group by 1
+having sum(p.amount) > (
+	select avg(customer_total)
+    from (
+		select sum(p.amount) as customer_total
+        from payment p
+        group by p.customer_id) as subquery
+     )
+     order by 2 desc;   
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