Assignment

1a. Display the first and last names of all actors from the table actor.
```
SELECT first_name, 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.
```
SELECT UPPER(CONCAT(first_name, ' ', last_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 first_name, last_name, actor_id
FROM actor
WHERE first_name = "Joe";
```

2b. Find all actors whose last name contain the letters GEN:

SELECT first_name, last_name, actor_id
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 first_name, last_name, actor_id
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 
```
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
ADD COLUMN middle_name varchar(40) AFTER first_name;
```
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 COLUMN middle_name;

4a. List the last names of actors, as well as how many actors have that last name.

SELECT last_name, count(last_name) AS 'last_name_frequency'
FROM actor
GROUP BY last_name
HAVING `last_name_frequency` >= 2;

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 'last_name_frequency'
FROM actor
GROUP BY last_name
Having `last_name_frequency` >= 2;
```
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 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 s.first_name, s.last_name, a.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 s.first_name, s.last_name, SUM(p.amount)
FROM staff AS s
INNER JOIN payment AS p
ON p.staff_id = s.staff_id
WHERE MONTH(p.payment_date) = 08 AND YEAR(p.payment_date) = 2005
GROUP BY s.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 f.title, COUNT(fa.actor_id) AS 'Actors'
FROM film_actor AS fa
INNER JOIN film as f
ON f.film_id = fa.film_id
GROUP BY f.title
ORDER BY Actors desc;

6d. How many copies of the film Hunchback Impossible exist in the inventory system?

SELECT title, COUNT(inventory_id) AS '# of copies'
FROM film
INNER JOIN inventory
USING (film_id)
WHERE title = 'Hunchback Impossible'
GROUP BY title;

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) AS 'Total Amount Paid'
FROM payment AS p 
JOIN customer AS c
ON p.customer_id = c.customer_id
GROUP BY c.customer_id
ORDER BY c.last_name;

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 'K%'
OR title LIKE 'Q%'
AND language_id IN
(
 SELECT language_id
 FROM language
 WHERE name = 'English'
);
```

7b. Use subqueries to display all actors who appear in the film Alone Trip.

SELECT first_name, last_name
FROM actor 
WHERE actor_id IN 
(
  SELECT actor_id
  FROM film_actor
  WHERE film_id = 
  (
     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.

SELECT first_name, last_name, email, country
FROM customer cus
JOIN address a
ON (cus.address_id = a.address_id)
JOIN city cit
ON (a.city_id = cit.city_id)
JOIN country ctr
ON (cit.country_id = ctr.country_id)
WHERE ctr.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 title, c.name
FROM film f
JOIN film_category fc
ON (f.film_id = fc.film_id)
JOIN category c
ON (c.category_id = fc.category_id)
WHERE name = 'family';

7e. Display the most frequently rented movies in descending order.

SELECT title, COUNT(title) as 'Rentals'
FROM film
JOIN inventory
ON (film.film_id = inventory.film_id)
JOIN rental
ON (inventory.inventory_id = rental.inventory_id)
GROUP by title
ORDER BY rentals desc;

7f. Write a query to display how much business, in dollars, each store brought in.

SELECT s.store_id, SUM(amount) AS Gross
FROM payment p
JOIN rental r
ON (p.rental_id = r.rental_id)
JOIN inventory i
ON (i.inventory_id = r.inventory_id)
JOIN store s
ON (s.store_id = i.store_id)
GROUP BY s.store_id;

7g. Write a query to display for each store its store ID, city, and country.

SELECT store_id, city, country
FROM store s
JOIN address a
ON (s.address_id = a.address_id)
JOIN city cit
ON (cit.city_id = a.city_id)
JOIN country ctr
ON(cit.country_id = ctr.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(amount) AS 'Total Sales', c.name AS 'Genre'
FROM payment p
JOIN rental r
ON (p.rental_id = r.rental_id)
JOIN inventory i
ON (r.inventory_id = i.inventory_id)
JOIN film_category fc
ON (i.film_id = fc.film_id)
JOIN category c
ON (fc.category_id = c.category_id)
GROUP BY c.name
ORDER BY SUM(amount) DESC;

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 top_five_genres AS
SELECT SUM(amount) AS 'Total Sales', c.name AS 'Genre'
FROM payment p
JOIN rental r
ON (p.rental_id = r.rental_id)
JOIN inventory i
ON (r.inventory_id = i.inventory_id)
JOIN film_category fc
ON (i.film_id = fc.film_id)
JOIN category c
ON (fc.category_id = c.category_id)
GROUP BY c.name
ORDER BY SUM(amount) DESC
LIMIT 5;

8b. How would you display the view that you created in 8a?
```
SELECT * 
FROM top_five_genres;
```
8c. You find that you no longer need the view top_five_genres. Write a query to delete it.
```
DROP VIEW top_five_genres;
```

Appendix: List of Tables in the Sakila DB

A schema is also available as sakila_schema.svg. Open it with a browser to view.

	'actor'
	'actor_info'
	'address'
	'category'
	'city'
	'country'
	'customer'
	'customer_list'
	'film'
	'film_actor'
	'film_category'
	'film_list'
	'film_text'
	'inventory'
	'language'
	'nicer_but_slower_film_list'
	'payment'
	'rental'
	'sales_by_film_category'
	'sales_by_store'
	'staff'
	'staff_list'
	'store'

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Appendix: List of Tables in the Sakila DB

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Assignment

Appendix: List of Tables in the Sakila DB