Solved Lab

lab-sql-joins.sql

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+-- Challenge - Joining on multiple tables
+-- Write SQL queries to perform the following tasks using the Sakila database:
+USE sakila;
+-- 1. List the number of films per category.
+SELECT c.name AS category_name, COUNT(fc.film_id) AS number_of_films 
+FROM category AS c
+LEFT JOIN film_category AS fc 
+ON c.category_id = fc.category_id
+GROUP BY c.name
+ORDER BY number_of_films DESC;
+
+-- 2. Retrieve the store ID, city, and country for each store.
+
+SELECT s.store_id, ci.city, co.country
+FROM store as s 
+LEFT JOIN address AS a ON s.address_id = a.address_id
+LEFT JOIN city as ci ON a.city_id = ci.city_id
+LEFT JOIN country as co ON ci.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 as s 
+LEFT JOIN staff as st ON s.store_id = st.store_id
+LEFT JOIN payment as p ON st.staff_id = p.staff_id
+GROUP BY s.store_id;
+
+-- 4. Determine the average running time of films for each category.
+SELECT cat.name, ROUND(AVG(f.length) , 2) AS avg_running_time
+FROM film as f 
+LEFT JOIN film_category as fc ON f.film_id = fc.film_id
+LEFT JOIN category as cat ON fc.category_id = cat.category_id
+GROUP BY cat.name
+ORDER BY avg_running_time DESC;
+
+-- Bonus:
+-- 5.Identify the film categories with the longest average running time.
+SELECT c.name, AVG(f.length) AS avg_length
+FROM category AS c 
+LEFT JOIN film_category AS fc ON c.category_id = fc.category_id
+LEFT JOIN film AS f ON fc.film_id = f.film_id
+GROUP BY c.name
+HAVING AVG(f.length) > (SELECT AVG(length) FROM film)
+ORDER BY avg_length DESC;
+
+-- 6. Display the top 10 most frequently rented movies in descending order.
+SELECT f.title, COUNT(r.rental_id) AS rental_count 
+FROM film AS F
+LEFT JOIN inventory AS i ON f.film_id = i.film_id
+LEFT JOIN rental AS r ON i.inventory_id = r.inventory_id
+GROUP BY f.title
+ORDER BY rental_count DESC
+LIMIT 10;
+
+-- 7. Determine if "Academy Dinosaur" can be rented from Store 1.
+SELECT film_id FROM film 
+WHERE title LIKE '%Academy Dinosaur%';
+
+SELECT * FROM inventory 
+WHERE film_id = (SELECT film_id FROM film 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 f.title,
+CASE 
+	WHEN MAX(i.inventory_id) IS NULL THEN 'NOT available'
+	ELSE 'Available'
+END AS availability_status
+FROM film AS f 
+LEFT JOIN inventory AS i on f.film_id = i.film_id
+GROUP BY f.title;