From f098e9603f966ce150d1669959e55b63ef7e467c Mon Sep 17 00:00:00 2001
From: Melissa Sanchez <melissa.sanchez.mx@gmail.com>
Date: Wed, 8 Jul 2026 18:58:02 +0200
Subject: [PATCH] lab solved

---
 sakila_subqueries.sql | 57 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 57 insertions(+)
 create mode 100644 sakila_subqueries.sql

diff --git a/sakila_subqueries.sql b/sakila_subqueries.sql
new file mode 100644
index 0000000..37692d1
--- /dev/null
+++ b/sakila_subqueries.sql
@@ -0,0 +1,57 @@
+USE sakila;
+
+-- 1. Determine the number of copies of the film "Hunchback Impossible" that exist in the inventory system
+SELECT * FROM inventory;
+SELECT * FROM film;
+
+-- step 1
+SELECT film_id
+FROM film
+WHERE title = "Hunchback Impossible";
+
+-- step 2
+SELECT COUNT(*)
+FROM inventory
+WHERE film_id = (
+SELECT film_id
+FROM film
+WHERE title = "Hunchback Impossible");
+
+-- 2. List all films whose length is longer than the average length of all the films in the Sakila database
+-- step 1
+SELECT AVG(length) AS avg_duration
+FROM film;
+-- step 2
+SELECT title 
+FROM film
+WHERE length > (
+SELECT AVG(length)
+FROM film);
+
+-- 3. Use a subquery to display all actors who appear in the film "Alone Trip"
+-- step 1
+SELECT film_id
+FROM film
+WHERE title = "Alone Trip";
+-- step 2
+SELECT actor_id
+FROM film_actor
+WHERE film_id = (
+SELECT film_id
+FROM film
+WHERE title = "Alone Trip");
+-- step 3
+SELECT first_name, last_name
+FROM actor
+WHERE actor_id IN (
+SELECT actor_id
+FROM film_actor
+WHERE film_id = (
+SELECT film_id
+FROM film
+WHERE title = "Alone Trip"));
+
+
+
+
+