SQueriosL
1.0.0
CREAR TABLA empleado ( empleado_id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, apellido VARCHAR(30) NOT NULL, nombre VARCHAR(30) NOT NULL, correo electrónico VARCHAR(100) NOT NULL, fecha_contratación FECHA NO NULL, notas MEDIUMTEXT, PRIMARY KEY (employee_id), INDEX (apellido), ÚNICO (correo electrónico));
MOTOR=InnoDB;
CREAR TABLA dirección ( empleado_id INTEGER UNSIGNED NOT NULL, dirección VARCHAR(50) NOT NULL, ciudad VARCHAR(30) NOT NULL, estado CHAR(2) NOT NULL, código postal CHAR(5) NOT NULL, CLAVE EXTRANJERA (employee_id) REFERENCIAS empleado ( empleado_id) )
MOTOR=InnoDB;
CREAR TABLA charset_example ( -> id INTEGER SIN FIRMAR NO NULO AUTO_INCREMENT, -> ascii_string VARCHAR(255) CONJUNTO DE CARACTERES ascii NO NULO, -> latin1_string VARCHAR(255) CONJUNTO DE CARACTERES latin1 NO NULO, -> utf8_string VARCHAR(255) CONJUNTO DE CARACTERES utf8 NO NULO, -> CLAVE PRIMARIA (id) -> );
INSERT INTO empleado(nombre, apellido, correo electrónico, fecha_contratación) VALORES ('Nischal', 'Bhatia', '[email protected]', '2014-12-15');
INSERT INTO empleado(nombre, apellido, correo electrónico, fecha_contratación) VALUES('Gurjot','Singh','[email protected]','2017-12-08');
INSERT INTO empleado(nombre, apellido, correo electrónico, fecha_contratación) VALUES('Jaskaran','Singh','[email protected]','2017-04-14');
INSERT INTO empleado(nombre, apellido, correo electrónico, fecha_contratación) VALORES ('Anjandeep', 'Singh', '[email protected]', '2017-11-30');
SELECCIONAR * DEL empleado;
INSERTAR EN dirección (id_empleado, dirección, ciudad, estado, código postal) VALORES (1, '123 Main Street', 'Cualquier ciudad', 'XE', '97052');
GROUP BY CLAUSE : :
- Suppose we want to find the top 10 customers who’ve spent the most money renting
movies.
SELECT customer_id, SUM(amount) AS Amt
FROM payment
-> GROUP BY customer_id
-> ORDER BY Amt DESC
-> LIMIT 10;
UPDATE CLAUSE : :
SELECT customer_id, first_name, last_name
FROM customer
WHERE first_name = 'Courtney' AND last_name = 'Day';
-- ACTUALIZAR cliente SET last_name = 'DAY-WEBB' WHERE customer_id = 245;
-- SELECCIONE id_cliente, nombre, apellido, última actualización DEL cliente DONDE nombre = 'Courtney';
DELETE CLAUSE : :
UPDATE customer
SET active = 0
WHERE customer_id = 245;
DELETE FROM rental / payment
WHERE customer_id = 245;
Now, we are allowed to delete the original P.K rows
DELETE FROM customer
WHERE customer_id = 245;
Retrieve all of the actors with “SON” in their last name and sort them alphabetically.
SELECT actor_id, last_name
FROM actor
WHERE last_name LIKE('%SON')
ORDER BY last_name ASC;
Calculate how many films there are for each rating category—G, PG, PG-13, R, and NC-17.
SELECT rating, COUNT(film_id) AS NumFilm
FROM film
GROUP BY rating;
What’s the ID of the customer who’s made the most visits to the video store?
SELECT customer_id, COUNT(payment_id) AS NumVisits, amount
FROM payment
GROUP BY customer_id
ORDER BY NumVisits DESC;
Let’s identify the top five actors who made the most film appearances and the
number of films they’ve each starred in.
SELECT actor_id, COUNT(actor_id) AS Appearances
FROM film_actor
GROUP BY actor_id
ORDER BY Appearances DESC
LIMIT 5;
Now, since actor_id is also present in the actor table
we use another SELECT statement to retrieve corresponding rows
SELECT actor_id, first_name, last_name
-> FROM actor
-> WHERE actor_id IN(107,102,198,181,23);
We have all the information we wanted, but unfortunately we still need to match
up the actors’ names and appearance counts manually
JOIN lets us do exactly that by connecting two or more tables based on a relationship that we specify.
SELECT a.first_name, a.last_name, COUNT(fa.actor_id) AS Appearance_Count
-> FROM film_actor fa JOIN actor a
-> ON a.actor_id = fa.actor_id
-> GROUP BY fa.actor_id
-> ORDER BY Appearance_Count DESC, a.first_name ASC, a.last_name ASC
-> LIMIT 5;
JOINS : : 3 Types : :
CREATE TABLE foo(foo_id INTEGER, foo_value CHAR(3));
CREATE TABLE bar(bar_id INTEGER, bar_value CHAR(3), foo_id INTEGER);
INSERT INTO foo(foo_id, foo_value)
-> VALUES(1, 'foo');
INSERT INTO foo(foo_id, foo_value)
-> VALUES(2, 'bar');
INSERT INTO bar(bar_id, bar_value, foo_id)
-> VALUES(1, 'baz', 2);
INSERT INTO bar(bar_id, bar_value, foo_id)
-> VALUES(2, 'qux', 3);
SHOW TABLES;
+---------------+
| Tables_in_foo |
+---------------+
| bar |
| foo |
+---------------+
SELECT *
-> FROM foo f INNER JOIN bar b
-> ON b.foo_id = f.foo_id;
+--------+-----------+--------+-----------+--------+
| foo_id | foo_value | bar_id | bar_value | foo_id |
+--------+-----------+--------+-----------+--------+
| 2 | bar | 1 | baz | 2 |
+--------+-----------+--------+-----------+--------+
mysql> SELECCIONAR * -> DESDE foo f BARRA DE UNIÓN EXTERNA IZQUIERDA b -> ENCENDIDO b.foo_id = f.foo_id;
+--------+-----------+--------+-----------+--------+
| foo_id | foo_value | bar_id | bar_value | foo_id |
+--------+-----------+--------+-----------+--------+
| 1 | foo | NULL | NULL | NULL |
| 2 | bar | 1 | baz | 2 |
+--------+-----------+--------+-----------+--------+
mysql> SELECCIONAR * -> DESDE foo f barra DE UNIÓN EXTERIOR DERECHA b -> ENCENDIDO b.foo_id = f.foo_id;
+--------+-----------+--------+-----------+--------+
| foo_id | foo_value | bar_id | bar_value | foo_id |
+--------+-----------+--------+-----------+--------+
| 2 | bar | 1 | baz | 2 |
| NULL | NULL | 2 | qux | 3 |
+--------+-----------+--------+-----------+--------+
mysql> SELECT * -> FROM foo JOIN bar;
+--------+-----------+--------+-----------+--------+
| foo_id | foo_value | bar_id | bar_value | foo_id |
+--------+-----------+--------+-----------+--------+
| 1 | foo | 1 | baz | 2 |
| 2 | bar | 1 | baz | 2 |
| 1 | foo | 2 | qux | 3 |
| 2 | bar | 2 | qux | 3 |
+--------+-----------+--------+-----------+--------+
ABSTRACTING WITH VIEWS : :
mysql> CREAR VISTA Actor_Appearance AS -> SELECCIONAR a.first_name, a.last_name, COUNT(fa.film_id) AS Appearance_Count -> FROM film_actor fa ÚNETE al actor a -> ON a.actor_id = fa.actor_id -> GROUP BY fa.actor_id ;
mysql> SELECCIONAR * -> DESDE Actor_Apariencia -> ORDENAR POR Apariencia_Count DESC -> LÍMITE 5;
+------------+-----------+------------------+
| first_name | last_name | Appearance_Count |
+------------+-----------+------------------+
| GINA | DEGENERES | 42 |
| WALTER | TORN | 41 |
| MARY | KEITEL | 40 |
| MATTHEW | CARREY | 39 |
| SANDRA | KILMER | 37 |
+------------+-----------+------------------+
### Duplication may arise, if ::
SELECT c.first_name, c.last_name, c.last_update, a.address, a.last_update
FROM customer AS c JOIN address AS a
ON a.address_id = c.address_id;
### Possible Solution is to provide alias or : :
mysql> CREAR VISTA cust_address(first_name, last_name, cust_last_update, dirección, add_last_update) AS -> SELECCIONAR c.first_name, c.last_name, c.last_update, a.address, a.last_update -> DEL cliente c UNIRSE a la dirección a -> ON a.address_id = c.address_id;
mysql> CREAR VER cliente_activo AS -> SELECCIONAR id_cliente, nombre, apellido, correo electrónico -> DEL cliente -> DONDE activo = 1;
NORMALIZING FORMS : :
THIRD NF :
SELECT a.actor_id, a.first_name, a.last_name, f.film_id, f.title
FROM film_actor fa
JOIN actor a ON fa.actor_id = a.actor_id
JOIN film f ON fa.film_id = f.film_id;
ALTERING TABLES : :
mysql> ALTER TABLE actor
-> ADD COLUMN bio VARCHAR(255) AFTER last_name;
mysql> ALTER TABLE actor
-> DROP COLUMN bio;
mysql> ALTER TABLE actor
-> ADD INDEX idx_last_update(last_update);
mysql> ALTER TABLE actor
-> DROP INDEX idx_last_update;
EXERCISE : :
mysql> SELECCIONE CONTAR (id_alquiler) COMO Num_Rented, id_cliente -> DESDE el alquiler -> GRUPO POR id_cliente -> PEDIR POR Num_Rented DESC -> LÍMITE 100;
Are there customers whose rental habits show they have a favorite actor?
Mire las otras tablas definidas en la base de datos sakila y observe cómo siguen 3NF.
SELECCIONE a.actor_id, a.first_name, a.last_name, f.film_id, f.title DE film_actor fa ÚNETE al actor a ON fa.actor_id = a.actor_id ÚNETE a la película f ON fa.film_id = f.film_id;
