--Abfrage 3: Alle Adressen mit voller Anschrift ausgeben
SELECT a.address_id, concat(a.address,", ",a.address2,", ",a.postal_code," ",c.city,", ",a.district,", ",co.country) AS "Full address"
FROM address AS a LEFT JOIN city AS c ON a.city_id = c.city_id LEFT JOIN country AS co ON c.country_id = co.country_id;
-- Abfrage 8: Welcher Kunde hat am meisten Umsatz generiert?
select concat(c.first_name," ",c.last_name) as "Kunde",sum(amount) AS "Umsatz" from payment AS r LEFT JOIN customer AS c ON c.customer_id = r.customer_id group by r.customer_id order by sum(amount) desc LIMIT 1;
-- Abfrage 9: Welche Einnahmen haben die Mitarbeiter gemacht?
select concat(s.name,", ",s.city) as "Mitarbeiter",sum(amount) AS "Einnahmen" from payment AS p LEFT JOIN staff_list AS s ON s.ID = p.staff_id group by p.staff_id order by sum(amount) desc;
-- Abfrage 10: zeigt an welcher Schauspieler in welchem Film mitspielt.
Select CONCAT(actor.first_name,' ', actor.last_name) as actorname, film.title
FROM sakila.actor, sakila.film, sakila.film_actor
WHERE (actor.actor_id = film_actor.actor_id)
AND (film.film_id = film_actor.film_id);
-- Abfrage 11: Gibt alle Filmtitel mit Kategorie und Rating aus
SELECT f.title AS "Titel", c.name AS "Kategorie", f.rating AS "Rating" FROM film_category AS fc LEFT JOIN category AS c ON fc.category_id = c.category_id LEFT JOIN film AS f ON fc.film_id = f.film_id;
-- Abfrage 15: Aus welcher Stadt kommen die meisten Kunden
Select city.city, count(customer_id) from city right join address on address.city_id = city.city_id right join customer on customer.address_id=address.address_id group by city.city_id order by count(customer_id) desc limit 1;