MySQL is a free and open-source relational database management system. It uses the SQL query language. I downloaded and used the Sakila MySQL database: A sample database that provides a standard schema for class activities, textbooks, etc. The Sakila database sports a highly integrated dataset, with many foreign keys pointing to many related tables. I used the MySQL Workbench to keep track of and execute my queries.
The exact operations are specified as comments above each of the code snippets below, and they included selecting, joining, filtering, and constructing views.
USE sakila;
-- 1. Display the first and last names of all actors from the table actor.
SELECT first_name, last_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
FROM actor
WHERE first_name = 'Joe';
-- 2b. Find all actors whose last name contain the letters GEN:
SELECT actor_id, first_name, last_name
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 actor_id, first_name, last_name
FROM actor
where last_name LIKE "%li%"
ORDER BY last_name ASC, first_name ASC;
-- 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. You want to keep a description of each actor. You don't think you will be performing queries on a description,
-- so create a column in the table actor named description and use the data type BLOB
-- (Make sure to research the type BLOB, as the difference between it and VARCHAR are significant).
ALTER TABLE actor
ADD COLUMN description BLOB AFTER last_name;
-- 3b. Very quickly you realize that entering descriptions for each actor is too much effort. Delete the description column.
ALTER TABLE actor
DROP COLUMN description;
-- 4a. List the last names of actors, as well as how many actors have that last name.
SELECT last_name, COUNT(last_name) as "Counts"
FROM actor
GROUP 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(last_name) as "Counts"
FROM actor
GROUP BY last_name
HAVING Counts >= 2;
-- 4c. The actor HARPO WILLIAMS was accidentally entered in the actor table as GROUCHO WILLIAMS. Write a query to fix the record.
UPDATE actor
set first_name = 'HARPO'
WHERE actor_id = 172;
-- 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.
UPDATE actor
set first_name = 'GROUCHO'
WHERE first_name = 'HARPO';
-- 5a. You cannot locate the schema of the address table. Which query would you use to re-create it?
SHOW CREATE TABLE 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 staff.first_name, staff.last_name, address.address
FROM staff
JOIN address ON staff.address_id=address.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 p.staff_id, concat(s.first_name, ' ', s.last_name) as Name, SUM(p.amount) as Total_Amount
FROM payment p
JOIN staff s ON p.staff_id = s.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 fa.film_id, f.title, count(fa.actor_id) as actor_count
FROM film_actor fa
INNER JOIN film f
ON (fa.film_id = f.film_id)
GROUP BY fa.film_id;
-- 6d. How many copies of the film Hunchback Impossible exist in the inventory system?
SELECT I.film_id, F.title, count(I.inventory_id) as count_copies
FROM inventory i, film f
WHERE i.film_id in
(SELECT f.film_id FROM film f
WHERE f.title like 'Hunchback%Impossible%')
AND i.film_id = f.film_id
GROUP BY 1
;
-- 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 c.first_name, c.last_name, sum(p.amount)
FROM customer c
JOIN payment p
ON c.customer_id = p.customer_id
GROUP BY c.last_name, c.first_name
ORDER BY c.last_name ASC
;
-- 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.
Select title
FROM film
WHERE (title LIKE "Q%" OR title LIKE "K%")
AND language_id = (
SELECT language_id
FROM language
WHERE name = 'English')
;
-- 7b. Use subqueries to display all actors who appear in the film Alone Trip.
select a.first_name, a.last_name
from film_actor fa, actor a
WHERE fa.film_id = (SELECT film_id FROM film f WHERE f.title = 'Alone Trip')
AND fa.actor_id = a.actor_id
;
-- 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.
SELECT first_name, last_name, email
FROM customer c
WHERE address_id in (SELECT a.address_id
FROM address a, city ci, country cn
where a.city_id = ci.city_id
AND ci.country_id = cn.country_id
AND cn.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 family films.
SELECT f.title
FROM film f
WHERE f.film_id in
(SELECT fc.film_id
FROM film_category fc, category c
WHERE fc.category_id = c.category_id
AND c.name = 'Family')
;
-- 7e. Display the most frequently rented movies in descending order.
SELECT f.title, count(*) as 'rentals'
FROM rental r, inventory i, film f
WHERE r.inventory_id = i.inventory_id
AND i.film_id = f.film_id
GROUP BY f.title
ORDER BY rentals DESC
;
-- 7f. Write a query to display how much business, in dollars, each store brought in.
SELECT c.store_id, SUM(p.amount) AS total_store
FROM payment p, customer c
WHERE p.customer_id = c.customer_id
GROUP BY c.store_id
;
-- 7g. Write a query to display for each store its store ID, city, and country.
SELECT s.store_id, ci.city, cn.country
FROM store s, address a, city ci, country cn
WHERE s.address_id = a.address_id
AND a.city_id = ci.city_id
AND ci.country_id = cn.country_id
;
-- 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.)
SELECT sum(pay.amount) as 'gross_revenue', cat.name as 'genre'
FROM category cat, film_category fcat, inventory inv, payment pay, rental rent
WHERE cat.category_id = fcat.category_id
AND rent.inventory_id = inv.inventory_id
AND pay.rental_id = rent.rental_id
AND inv.film_id = fcat.film_id
GROUP BY genre;
-- 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.
CREATE VIEW Top5Genres3 AS
SELECT sum(pay.amount) as 'gross_revenue', cat.name as 'genre'
FROM category cat, film_category fcat, inventory inv, payment pay, rental rent
WHERE cat.category_id = fcat.category_id
AND rent.inventory_id = inv.inventory_id
AND pay.rental_id = rent.rental_id
AND inv.film_id = fcat.film_id
GROUP BY genre
ORDER BY gross_revenue DESC
LIMIT 5;
-- 8b. How would you display the view that you created in 8a?
SELECT * FROM Top5Genres3;
-- 8c. You find that you no longer need the view top_five_genres. Write a query to delete it.
DROP VIEW Top5Genres3;