Soluciones sql

insert into actor (first_name, last_name)
values ('juan','perez');

insert into actor (first_name, last_name)
values ('rosa','pi');

update actor set first_name='PEPE'
where actor_id=10;

update actor set first_name=concat('Actor',actor_id)
where actor_id>200;

delete from actor where actor_id>200;

select country, customer.* from country join city on country.country_id=city.country_id
join address on city.city_id=address.city_id
join customer on address.address_id=customer.address_id
where country like 'A%';

select country.country, customer.* from customer join address on customer.address_id=address.address_id
join city on address.city_id=city.city_id
join country on city.country_id=country.country_id
where country like 'a%';

select distinct first_name,last_name from actor join film_actor on actor.actor_id=film_actor.actor_id
join film on film_actor.film_id=film.film_id
where length>140;

select distinct name from category join film_category on category.category_id=film_category.category_id
join film on film_category.film_id=film.film_id
where rating='r';

select title, count(rental_id) total from film join inventory on film.film_id=inventory.film_id
join rental on inventory.inventory_id=rental.inventory_id
group by film.film_id
having total>20;

select country.country, customer.* from customer join address on customer.address_id=address.address_id
join city on address.city_id=city.city_id
join country on city.country_id=country.country_id
where country ='spain' or country='argentina';

select country.country, customer.* from customer join address on customer.address_id=address.address_id
join city on address.city_id=city.city_id
join country on city.country_id=country.country_id
where country in ('spain','argentina');

select title, name from category join film_category on category.category_id=film_category.category_id
join film on film_category.film_id=film.film_id
where name in ('children','family');

select first_name, last_name from actor
where first_name like '%x%' or last_name like '%x%';

select * from address where district='california' and phone like '%274%';

Publicado por

Juan Pablo Fuentes

Formador de programación y bases de datos