Sakila_Homework.sql

# Robert Orr's thing

-- 1a. You need a list of all the actors who have Display the first and last names of all actors from the table actor. 
-- 1b. Display the first and last name of each actor in a single column in upper case letters. Name the column Actor Name. 
-- 2a. You need to find the ID number, first name, and last name of an actor, of whom you know only the first name, "Joe." What is one query would you use to obtain this information?
-- 2b. Find all actors whose last name contain the letters GEN:
-- 2c. Find all actors whose last names contain the letters LI. This time, order the rows by last name and first name, in that order:
-- 2d. Using IN, display the country_id and country columns of the following countries: Afghanistan, Bangladesh, and China:
-- 3a. Add a middle_name column to the table actor. Position it between first_name and last_name. Hint: you will need to specify the data type.
-- 3b. You realize that some of these actors have tremendously long last names. Change the data type of the middle_name column to blobs.
-- 3c. Now delete the middle_name column.
-- 4a. List the last names of actors, as well as how many actors have that last name.
-- 4b. List last names of actors and the number of actors who have that last name, but only for names that are shared by at least two actors
-- 4c. Oh, no! The actor HARPO WILLIAMS was accidentally entered in the actor table as GROUCHO WILLIAMS, the name of Harpo's second cousin's husband's yoga teacher. Write a query to fix the record.
-- 4d. Perhaps we were too hasty in changing GROUCHO to HARPO. It turns out that GROUCHO was the correct name after all! In a single query, if the first name of the actor is currently HARPO, change it to GROUCHO. Otherwise, change the first name to MUCHO GROUCHO, as that is exactly what the actor will be with the grievous error. BE CAREFUL NOT TO CHANGE THE FIRST NAME OF EVERY ACTOR TO MUCHO GROUCHO, HOWEVER! (Hint: update the record using a unique identifier.)
-- 5a. You cannot locate the schema of the address table. Which query would you use to re-create it?
-- 6a. Use JOIN to display the first and last names, as well as the address, of each staff member. Use the tables staff and address:
-- 6b. Use JOIN to display the total amount rung up by each staff member in August of 2005. Use tables staff and payment. 
-- 6c. List each film and the number of actors who are listed for that film. Use tables film_actor and film. Use inner join.
-- 6d. How many copies of the film Hunchback Impossible exist in the inventory system?
-- 6e. Using the tables payment and customer and the JOIN command, list the total paid by each customer. List the customers alphabetically by last name:
-- 
-- 
--     ![Total amount paid](Images/total_payment.png)
-- 
-- 7a. The music of Queen and Kris Kristofferson have seen an unlikely resurgence. As an unintended consequence, films starting with the letters K and Q have also soared in popularity. Use subqueries to display the titles of movies starting with the letters K and Q whose language is English. 
-- 7b. Use subqueries to display all actors who appear in the film Alone Trip.
-- 7c. You want to run an email marketing campaign in Canada, for which you will need the names and email addresses of all Canadian customers. Use joins to retrieve this information.
-- 7d. Sales have been lagging among young families, and you wish to target all family movies for a promotion. Identify all movies categorized as famiy films.
-- 7e. Display the most frequently rented movies in descending order.
-- 7f. Write a query to display how much business, in dollars, each store brought in.
-- 7g. Write a query to display for each store its store ID, city, and country.
-- 7h. List the top five genres in gross revenue in descending order. (Hint: you may need to use the following tables: category, film_category, inventory, payment, and rental.)
-- 8a. In your new role as an executive, you would like to have an easy way of viewing the Top five genres by gross revenue. Use the solution from the problem above to create a view. If you haven't solved 7h, you can substitute another query to create a view.
-- 8b. How would you display the view that you created in 8a?
-- 8c. You find that you no longer need the view top_five_genres. Write a query to delete it.
-- 

use sakila;
-- 1a. You need a list of all the actors who have Display the first and last names of all actors from the table actor. 
SELECT actor.first_name, actor.last_name
FROM actor;
-- 1b. Display the first and last name of each actor in a single column in upper case letters. Name the column Actor Name. 
ALTER TABLE actor DROP actor_name;
ALTER TABLE actor ADD COLUMN actor_name VARCHAR(50);
SELECT actor_id, CONCAT(first_name, " ",last_name) AS actor_name FROM actor;
SELECT actor_id Actor_Name, UPPER(Actor_Name) AS Actor_Name FROM actor;
-- 2a. You need to find the ID number, first name, and last name of an actor, of whom you know only the first name, "Joe." What is one query would you use to obtain this information?
SELECT actor_id, first_name, last_name, actor_name
FROM actor
WHERE first_name IN ('Joe');
-- 2b. Find all actors whose last name contain the letters GEN:
select * 
from actor 
where last_name like '%GEN%';
-- 2c. Find all actors whose last names contain the letters LI. This time, order the rows by last name and first name, in that order:
select * 
from actor 
where last_name like '%LI%'
ORDER BY last_name, first_name;
-- 2d. Using IN, display the country_id and country columns of the following countries: Afghanistan, Bangladesh, and China:
SELECT country_id, country
FROM country
WHERE country IN ('Afghanistan', 'Bangladesh', 'China');

-- 3a. Add a middle_name column to the table actor. Position it between first_name and last_name. Hint: you will need to specify the data type.
#ALTER TABLE actor DROP middle_name;
ALTER TABLE actor ADD COLUMN middle_name VARCHAR(50);
select first_name, middle_name, last_name from actor;
-- 3b. You realize that some of these actors have tremendously long last names. Change the data type of the middle_name column to blobs.
ALTER TABLE actor MODIFY COLUMN middle_name BLOB;
-- 3c. Now delete the middle_name column.
ALTER TABLE actor DROP middle_name;
-- 4a. List the last names of actors, as well as how many actors have that last name.
select last_name, count(*)
  from actor
  group by last_name
  ORDER BY last_name
  ;
-- 4b. List last names of actors and the number of actors who have that last name, but only for names that are shared by at least two actors
select last_name, count(*)
  from actor
  group by last_name
  HAVING count(*) > 1
  ;
-- 4c. Oh, no! The actor HARPO WILLIAMS was accidentally entered in the actor table as GROUCHO WILLIAMS, the name of Harpo's second cousin's husband's yoga teacher. 
	# Write a query to fix the record.
UPDATE actor
SET first_name = 'HARPO'
WHERE first_name = 'GROUCHO' AND last_name = 'WILLIAMS';
-- 4d. Perhaps we were too hasty in changing GROUCHO to HARPO. It turns out that GROUCHO was the correct name after all! In a single query, 
-- 	if the first name of the actor is currently HARPO, change it to GROUCHO. 
-- 	Otherwise, change the first name to MUCHO GROUCHO, as that is exactly what the actor will be with the grievous error.
-- 	BE CAREFUL NOT TO CHANGE THE FIRST NAME OF EVERY ACTOR TO MUCHO GROUCHO, HOWEVER! (Hint: update the record using a unique identifier.)
UPDATE actor
SET first_name =
CASE
  WHEN first_name = 'HARPO'
  THEN 'GROUCHO'
ELSE 'MUCHO GROUCHO'
END
WHERE actor_id = 172;

-- 5a. You cannot locate the schema of the address table. Which query would you use to re-create it?
SHOW CREATE TABLE sakila.address;

-- 6a. Use JOIN to display the first and last names, as well as the address, of each staff member. Use the tables staff and address:
SELECT first_name, last_name, address
FROM staff s
INNER JOIN address a
ON s.address_id = a.address_id;

-- 6b. Use JOIN to display the total amount rung up by each staff member in August of 2005. Use tables staff and payment. 
SELECT first_name, last_name, SUM(amount)
FROM staff s
INNER JOIN payment p
ON s.staff_id = p.staff_id
GROUP BY p.staff_id;

-- 6c. List each film and the number of actors who are listed for that film. Use tables film_actor and film. Use inner join.
SELECT title, COUNT(actor_id)
FROM film f
INNER JOIN film_actor fa
ON f.film_id = fa.film_id
GROUP BY title;

-- 6d. How many copies of the film Hunchback Impossible exist in the inventory system?
SELECT title, COUNT(inventory_id)
FROM film f
INNER JOIN inventory i 
ON f.film_id = i.film_id
WHERE title = 'Hunchback Impossible';


-- 6e. Using the tables payment and customer and the JOIN command, list the total paid by each customer. List the customers alphabetically by last name:
SELECT last_name, first_name, SUM(amount)
FROM payment p
INNER JOIN customer c
ON p.customer_id = c.customer_id
GROUP BY p.customer_id
ORDER BY last_name ASC;

--     ![Total amount paid](Images/total_payment.png)

-- 7a. The music of Queen and Kris Kristofferson have seen an unlikely resurgence. 
-- As an unintended consequence, films starting with the letters K and Q have also soared in popularity. 
-- Use subqueries to display the titles of movies starting with the letters K and Q whose language is English.
USE Sakila;
SELECT title FROM film
WHERE language_id in
	(SELECT language_id 
	FROM language
	WHERE name = 'English' )
AND (title LIKE 'K%') OR (title LIKE 'Q%');
-- 7b. Use subqueries to display all actors who appear in the film Alone Trip.
USE Sakila;
SELECT last_name, first_name
FROM actor
WHERE actor_id in
	(SELECT actor_id FROM film_actor
	WHERE film_id in 
		(SELECT film_id FROM film
		WHERE title = "Alone Trip"));
-- 7c. You want to run an email marketing campaign in Canada, for which you will need the names and email addresses of all Canadian customers. Use joins to retrieve this information.
USE Sakila;
SELECT country, last_name, first_name, email
FROM country c
LEFT JOIN customer cu
ON c.country_id = cu.customer_id
WHERE country = 'Canada';
-- 7d. Sales have been lagging among young families, and you wish to target all family movies for a promotion. Identify all movies categorized as famiy films.
USE Sakila;
SELECT title, category
FROM film_list
WHERE category = 'Family';
-- 7e. Display the most frequently rented movies in descending order.
USE Sakila;
SELECT i.film_id, f.title, COUNT(r.inventory_id)
FROM inventory i
INNER JOIN rental r
ON i.inventory_id = r.inventory_id
INNER JOIN film_text f 
ON i.film_id = f.film_id
GROUP BY r.inventory_id
ORDER BY COUNT(r.inventory_id) DESC;
-- 7f. Write a query to display how much business, in dollars, each store brought in.
SELECT store.store_id, SUM(amount)
FROM store
INNER JOIN staff
ON store.store_id = staff.store_id
INNER JOIN payment p 
ON p.staff_id = staff.staff_id
GROUP BY store.store_id
ORDER BY SUM(amount);
-- 7g. Write a query to display for each store its store ID, city, and country.
USE Sakila;
SELECT s.store_id, city, country
FROM store s
INNER JOIN customer cu
ON s.store_id = cu.store_id
INNER JOIN staff st
ON s.store_id = st.store_id
INNER JOIN address a
ON cu.address_id = a.address_id
INNER JOIN city ci
ON a.city_id = ci.city_id
INNER JOIN country coun
ON ci.country_id = coun.country_id;
-- 7h. List the top five FinalFive in gross revenue in descending order. (Hint: you may need to use the following tables: category, film_category, inventory, payment, and rental.)
SELECT c.name, SUM(p.amount) AS Gross_Revenue
FROM category c
JOIN film_category fc ON (c.category_id=fc.category_id)
JOIN inventory i ON (fc.film_id=i.film_id)
JOIN rental r ON (i.inventory_id=r.inventory_id)
JOIN payment p ON (r.rental_id=p.rental_id)
GROUP BY c.name ORDER BY Gross_Revenue  LIMIT 5;

-- 8a. In your new role as an executive, you would like to have an easy way of viewing the Top five genres by gross revenue. Use the solution from the problem above to create a view. 
-- If you haven't solved 7h, you can substitute another query to create a view.
DROP VIEW IF EXISTS FinalFive;
CREATE VIEW FinalFive AS
SELECT c.name, SUM(p.amount) AS Gross_Revenue 
FROM category c
JOIN film_category fc ON (c.category_id=fc.category_id)
JOIN inventory i ON (fc.film_id=i.film_id)
JOIN rental r ON (i.inventory_id=r.inventory_id)
JOIN payment p ON (r.rental_id=p.rental_id)
GROUP BY c.name ORDER BY Gross_Revenue  LIMIT 5;
-- -- 8b. How would you display the view that you created in 8a?
SELECT * FROM FinalFive;
-- 8c. You find that you no longer need the view top_five_genres. Write a query to delete it.
DROP VIEW FinalFive;




-- select a.name, b.name, a.city
-- from customer_list as a
-- join customer_list as b
-- using (city)
-- where a.id <> b.id

-- SELECT customer.first_name, customer.last_name, address.address
-- FROM customer
-- 
-- create index my_custom_index
-- on film(replacement_cost)
-- 
-- show index
-- from film;