Junior - lab select done

select.sql

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+USE sakila;
+
+-- 1. How many distinct (different) actors' last names are there?
+
+SELECT COUNT(DISTINCT(last_name)) FROM actor;
+
+-- 2. In how many different languages where the films originally produced? (Use the column language_id from the film table)
+
+SELECT COUNT(DISTINCT(language_id)) FROM film;
+
+-- 3. How many movies were released with "PG-13" rating?
+
+SELECT COUNT(rating) from film
+WHERE rating = "PG-13";
+
+-- 4. Get 10 the longest movies from 2006.
+
+SELECT * from film
+where release_year >= 2006
+ORDER BY length DESC limit 10;
+
+-- 5. How many days has been the company operating (check DATEDIFF() function)?
+
+SELECT DATEDIFF(max(return_date), min(rental_date)) from rental;
+
+-- 6. Show rental info with additional columns month and weekday. Get 20.
+
+SELECT
+	rental_date,
+	month(rental_date) as month,
+	DAYOFWEEK(rental_date) as weekday
+from rental
+limit 20;
+
+-- 7. Add an additional column day_type with values 'weekend' and 'workday' depending on the rental day of the week.
+
+
+set sql_safe_updates =0;
+ALTER TABLE rental
+ADD day_type VARCHAR(40);
+
+
+UPDATE rental
+SET day_type = CASE
+    WHEN DAYOFWEEK(rental_date) IN (1, 7) THEN 'weekend'
+    ELSE 'workday'
+END;
+
+-- 8 How many rentals were in the last month of activity?
+
+SELECT COUNT(*)
+FROM rental
+WHERE MONTH(rental_date) = (
+    SELECT MONTH(MAX(rental_date)) 
+    FROM rental
+);
+
+-- 9 Get film ratings.
+
+SELECT title, rating from film
+
+-- 10 Get release years.
+
+SELECT title, release_year from film
+
+-- 11 Get all films with ARMAGEDDON in the title.
+
+SELECT title from film
+where title LIKE "%ARMAGEDDON%"
+
+-- 12 Get all films with APOLLO in the title
+
+SELECT title from film
+where title LIKE "%APOLLO%"
+
+
+-- 13 Get all films which title ends with APOLLO.
+
+SELECT title from film
+where title LIKE "%APOLLO"
+
+-- 14 Get all films with word DATE in the title.
+
+SELECT title from film
+WHERE title REGEXP '\\bDATE\\b';
+
+-- 15 Get 10 films with the longest title.
+
+SELECT title
+FROM film
+ORDER BY LENGTH(title) DESC
+LIMIT 10;
+
+-- 16 Get 10 the longest films.
+
+SELECT title, length
+FROM film
+ORDER BY LENGTH(length) DESC
+LIMIT 10;
+
+-- 17 How many films include Behind the Scenes content?
+
+SELECT COUNT(*)
+FROM film
+WHERE special_features LIKE "%Behind the Scenes%";
+
+
+-- 18 List films ordered by release year and title in alphabetical order.
+
+SELECT release_year, title FROM film
+ORDER BY release_year 
+
+-- 19 Drop column picture from staff.
+
+ALTER TABLE staff
+DROP column picture;
+
+-- 20 A new person is hired to help Jon. Her name is TAMMY SANDERS, and she is a customer.
+-- Update the database accordingly.
+
+INSERT INTO staff VALUES("3", "TAMMY", "SANDERS", 4, "tammy.sandres@sakilastaff.com", "2", "1", "Tammy", "tammy", "2023-10-25");
+
+
+-- 21 Add rental for movie "Academy Dinosaur" by Charlotte Hunter from Mike Hillyer at Store 1.
+-- You can use current date for the rental_date column in the rental table.
+-- Hint: Check the columns in the table rental and see what information you would need to add there.
+-- You can query those pieces of information. For eg., you would notice that you need customer_id information as well.
+-- To get that you can use the following query:
+
+-- select customer_id from sakila.customer
+-- where first_name = 'CHARLOTTE' and last_name = 'HUNTER';
+-- Use similar method to get inventory_id, film_id, and staff_id.
+
+SELECT customer_id
+FROM customer
+WHERE first_name = 'CHARLOTTE' AND last_name = 'HUNTER';
+
+SELECT film_id
+FROM film
+WHERE title = 'Academy Dinosaur';
+
+SELECT staff_id
+FROM staff
+WHERE first_name = 'Mike' AND last_name = 'Hillyer';
+
+SELECT i.inventory_id
+FROM inventory AS i
+JOIN film AS f ON i.film_id = f.film_id
+JOIN store AS s ON i.store_id = s.store_id
+WHERE f.title = 'Academy Dinosaur' AND s.store_id = 1;
+
+INSERT INTO rental (rental_date, inventory_id, customer_id, return_date, staff_id)
+VALUES (CURRENT_DATE, inventory_id, customer_id, NULL, staff_id);
+
+
+
+-- 22 Delete non-active users, but first, create a backup table deleted_users to store customer_id, email, and the date for the users that would be deleted. Follow these steps:
+
+-- Check if there are any non-active users
+SELECT count(active) FROM customer WHERE active = 0 -- yes there are! 15 
+​
+-- Create a table backup table as suggested
+CREATE TABLE back_up_table AS SELECT * FROM customer where active = 0;
+SELECT * FROM back_up_table;
+
+-- Insert the non active users in the table backup table
+-- I'm actually just str8 up copying if it's non-active which is faster (?)
+
+-- Delete the non active users from the table customer
+SET FOREIGN_KEY_CHECKS=0; -- because: #1451 - Cannot delete or update a parent row: a foreign key constraint fails 
+-- (...) , CONSTRAINT ... FOREIGN KEY 
+-- (...) REFERENCES ... (..))
+
+DELETE FROM customer
+WHERE active = 0;
+
+SET FOREIGN_KEY_CHECKS=1;
+
+select * from customer;