Solved lab

lab_joins.sql

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+USE sakila;
+
+-- ## Challenge - Joining on multiple tables
+
+-- Write SQL queries to perform the following tasks using the Sakila database:
+
+
+-- 1. List the number of films per category.
+SELECT c.name AS category, COUNT(*) FROM film f
+JOIN film_category fc
+ON f.film_id = fc.film_id
+JOIN category c
+ON c.category_id = fc.category_id
+GROUP BY category;
+
+-- 2. Retrieve the store ID, city, and country for each store.
+SELECT s.store_id, c.city, co.country
+FROM store s
+JOIN address a ON s.address_id = a.address_id
+JOIN city c ON a.city_id = c.city_id
+JOIN country co ON c.country_id = co.country_id;
+
+
+-- 3.  Calculate the total revenue generated by each store in dollars.
+SELECT s.store_id, SUM(p.amount) AS total_revenue
+FROM store s
+JOIN staff sf ON s.store_id = sf.store_id
+JOIN payment p ON p.staff_id = sf.staff_id
+GROUP BY s.store_id;
+
+
+-- .  Determine the average running time of films for each category.
+SELECT c.name AS category, AVG(f.length) AS avg_running_time
+FROM film f
+JOIN film_category fc ON f.film_id = fc.film_id
+JOIN category c ON fc.category_id = c.category_id
+GROUP BY category;
+
+
+-- **Bonus**:
+
+-- 5.  Identify the film categories with the longest average running time.,
+SELECT c.name AS category, AVG(f.length) AS avg_running_time
+FROM film f
+JOIN film_category fc ON f.film_id = fc.film_id
+JOIN category c ON fc.category_id = c.category_id
+GROUP BY category
+ORDER BY avg_running_time DESC;
+
+-- 6.  Display the top 10 most frequently rented movies in descending order.
+SELECT f.title AS Movie, COUNT(*) AS rented_movies
+FROM rental r
+JOIN inventory i ON r.inventory_id = i.inventory_id
+JOIN film f ON i.film_id = f.film_id
+GROUP BY f.title, f.film_id
+ORDER BY rented_movies DESC
+LIMIT 10;
+
+-- 7. Determine if "Academy Dinosaur" can be rented from Store 1.
+SELECT title, f.film_id, store_id FROM film f
+JOIN inventory i ON f.film_id = i.film_id
+WHERE title = "Academy Dinosaur" AND store_id = 1; 
+
+-- 8. Provide a list of all distinct film titles, along with their availability status in the inventory. 
+-- Include a column indicating whether each title is 'Available' or 'NOT available.' Note that there are 42 titles 
+-- that are not in the inventory, and this information can be obtained using a `CASE` statement combined with `IFNULL`."
+SELECT DISTINCT f.title,
+CASE 
+WHEN i.film_id IS NULL THEN "Not_available"
+ELSE "Avaialable"
+END AS Availability_status
+FROM film f
+JOIN inventory i ON f.film_id = i.film_id
+GROUP BY f.film_id, f.title;
+