Solved lab

solutions.sql

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+USE sakila;
+SHOW TABLES;
+SELECT * FROM actor;
+SELECT * FROM film;
+SELECT * FROM customer;
+SELECT title FROM film;
+SELECT name AS language
+FROM language;
+SELECT first_name
+FROM staff;
+SELECT DISTINCT release_year
+FROM film
+ORDER BY release_year;
+SELECT COUNT(*) AS store_count
+FROM store;
+SELECT COUNT(*) AS employee_count
+FROM staff;
+-- Distinct films that exist in inventory (available for rent)
+SELECT COUNT(DISTINCT film_id) AS films_available_for_rent
+FROM inventory;
+
+-- Distinct films that appear in rentals (have been rented at least once)
+SELECT COUNT(DISTINCT i.film_id) AS films_that_have_been_rented
+FROM rental r
+JOIN inventory i ON r.inventory_id = i.inventory_id;
+SELECT COUNT(DISTINCT last_name) AS distinct_actor_last_names
+FROM actor;
+SELECT film_id, title, length
+FROM film
+ORDER BY length DESC, title ASC
+LIMIT 10;
+SELECT actor_id, first_name, last_name
+FROM actor
+WHERE first_name = 'SCARLETT';
+SELECT film_id, title, length
+FROM film
+WHERE title LIKE '%ARMAGEDDON%'
+  AND length > 100;
+-- special_features is a SET; FIND_IN_SET ensures exact match within the list
+SELECT COUNT(*) AS films_with_behind_the_scenes
+FROM film
+WHERE FIND_IN_SET('Behind the Scenes', special_features) > 0;