Ejercicios PHP

Tenemos una variable $tam y queremos una lista (ul/li) con la longitud de $tam.

Ejemplo: $tam=4 nos muestra:

  • Elemento 1
  • Elemento 2
  • Elemento 3
  • Elemento 4

Tenemos una variable $long y queremos un árbol de asteriscos de longitud $long

Ejemplo: $long=6 nos muestra:

*
**
***
****
*****
******

Lo mismo con un rombo:

  **
 ****
******
******
 **** 
  **

Tenemos una variable $lado y queremos una tabla cuadrada de ese lado.
Ejemplo: $lado = 4

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

Más pasitos en PHP

 <?php
 $a = 3;
 $b = 3;
 if ($a > $b * 2) {
 echo 'a es mucho mayor que b';
 } elseif ($a > $b) {
 echo 'a es mayor que b';
 } elseif ($a * 2 < $b) {
 echo 'a es mucho menor que b';
 } elseif ($a < $b) {
 echo 'a es menor que b';
 } else {
 echo 'iguales';
 }

 echo 'hola que tal';
 if ($a) {
 
 }
 ?> 
 <hr/>
 <?php
 $a = 1;
 while ($a ** 3 < 1000) {
 echo ($a ** 3) . "<br/>";
 $a++;
 }
 ?>
 <hr/>
 <?php
 for ($i = 1; $i < 100; $i*=2) {
 echo $i . "<br/>";
 }

 $tabla = array('Ana', 'Juan', 'Rosa');
 foreach ($tabla as $nombre) {
 echo $nombre . "<br/>";
 }
 ?>
 <hr/>
 <?php
 $tabla = 5;
 //Imprimiendo el HTML
 echo "<table border=1>";
 for ($i = 1; $i <= 10; $i++) {
 echo "<tr><td>$i</td><td>x</td><td>$tabla</td><td>" . ($i * $tabla) . "</td></tr>";
 }
 echo "</table>";
 
 //Mezclando HTML y PHP. Recordemos que <?= es equivalente a <?php echo 
 ?>
 <table border=1>
 <?php
 for ($i = 1; $i <= 10; $i++) {
 ?>
 <tr><td><?= $i ?></td><td>x</td><td><?= $tabla ?></td><td><?= ($i * $tabla) ?></td></tr>
 <?php
 }
 ?>
 </table>

Primeros pasos en PHP

 <?php
 $a=2;
 $b=3;
 echo "suma: \n";
 echo $a+$b; //Muestra 5
 echo "<br/>concatenar: ";
 echo $a.$b; //Muestra 23
 ?>
 <hr/>
 
 <?php
 $a=5;
 $b=4.2;
 $d=false;
 $e="7up";
 echo $a+$e;
 echo gettype($a);
 echo "<br/>";
 echo gettype($b);
 echo "<br/>";
 echo gettype($d);
 echo "<br/>";
 echo gettype($e);
 echo "<br/>";
 ?>
 <hr/>
 <?php 
 $a=5;
 $b=true;
 echo "#".($a==$b)."#";
 echo "<br/>";
 echo "#".($a===$b)."#";
 ?>

Crear funciones

Ejemplos:

CREATE DEFINER=`root`@`localhost` 
FUNCTION `inventory_in_stock`(p_inventory_id INT)
 RETURNS tinyint(1)
 READS SQL DATA
BEGIN
 DECLARE v_rentals INT;
 DECLARE v_out INT;
 SELECT COUNT(*) INTO v_rentals
 FROM rental
 WHERE inventory_id = p_inventory_id;

 IF v_rentals = 0 THEN
 RETURN TRUE;
 END IF;

 SELECT COUNT(rental_id) INTO v_out
 FROM inventory LEFT JOIN rental USING(inventory_id)
 WHERE inventory.inventory_id = p_inventory_id
 AND rental.return_date IS NULL;

 IF v_out > 0 THEN
 RETURN FALSE;
 ELSE
 RETURN TRUE;
 END IF;
END

select inventory_in_stock(30)

CREATE DEFINER=`root`@`localhost`
 FUNCTION `clientes_por_pais`
 (p_country varchar(50)) RETURNS int(11)
BEGIN
declare total int;

SELECT count(customer_id) into total
 FROM sakila.country
left join city using (country_id)
left join address using (city_id)
left join customer using (address_id)
where country=p_country
group by country;

RETURN total;
END

select clientes_por_pais('Algeria')

CREATE DEFINER=`root`@`localhost` FUNCTION `peliculas_por_actor`(p_actor_id int) RETURNS int(11)
BEGIN

declare total int;

select count(film_id) into total from film join
film_actor using (film_id)
join actor using(actor_id)
where actor.actor_id=p_actor_id;

RETURN total;
END

select peliculas_por_actor(1)

CREATE DEFINER=`root`@`localhost` 
FUNCTION `total_ventas`() RETURNS decimal(10,2)
BEGIN
declare total decimal(10,2);
select sum(amount) into total from payment;
RETURN total;
END

Sql views

create or replace view cliente_gaston as
select customer.customer_id first_name, last_name, sum(amount) as total
from customer join payment using(customer_id)
group by first_name, last_name
order by total desc
limit 0,5
update customer set active=5
where customer_id in (select customer_id from cliente_gaston)

Más SQL

Países con tiendas:

select country, count(store_id) as total
from country left join city using (country_id)
left join address using (city_id)
left join store using (address_id)
group by country
having total>0

Cliente que más gasta:

select first_name, last_name, sum(amount) as total
from customer join payment using(customer_id)
group by first_name, last_name
order by total desc
limit 0,1

Ejemplos de CASE e IF:

SELECT *,
case when amount<1 then 'barato'
when amount between 1 and 3 then 'medio'
else 'caro' end precio,
if (amount<3,'barato','caro') precio2
FROM sakila.payment;

Con funciones de agregado:

select first_name, last_name, sum(amount) as total,
if (sum(amount)>100,'gastador','rácano') tipo
from customer join payment using(customer_id)
group by customer_id

Pagos formateados:

SELECT lpad(format(amount,3),10,' ') from payment

Nombre formateado y ordenar por longitud de apellido:

select ucase(first_name), lcase(last_name),
concat(
ucase(substring(first_name,1,1)),
lcase(substring(first_name,2))
) cliente,
length(last_name) longitud
from customer
order by longitud desc

Sentencias sql

Clientes que se llaman igual que actores de películas

SELECT distinct concat(c.first_name, " ", c.last_name) cliente, 
concat(a.first_name, " ", a.last_name) as actor
FROM actor a join
film_actor using (actor_id)
join film using (film_id)
join inventory using (film_id)
join rental using (inventory_id)
join customer c using (customer_id)
where a.first_name=c.first_name
order by cliente

Películas por categoría:

select name, count(title)
from category join film_category using(category_id)
join film using (film_id)
group by name

Clientes por país:

SELECT country, count(customer_id) FROM sakila.country
left join city using (country_id)
left join address using (city_id)
left join customer using (address_id)
group by country;

Clientes que se apellidan como el empleado que los atiende:

select distinct concat(c.first_name, " ", c.last_name) cliente, 
concat(s.first_name, " ", s.last_name) as empleado
from customer c join rental using (customer_id)
join staff s using (staff_id)
where c.last_name=s.last_name

Clientes que viven en el mismo código postal que los empleados:

select distinct concat(c.first_name, " ", c.last_name) cliente, 
concat(s.first_name, " ", s.last_name) as empleado
from customer c join address a1 on c.address_id=a1.address_id
join address a2 on a1.postal_code=a2.postal_code
join staff s on a2.address_id=s.address_id

Películas con importe total de alquiler menor que 2:

select title, avg(ifnull(amount,0)) total
from film left join inventory using (film_id)
left join rental using (inventory_id)
left join payment using (rental_id)
group by title
having total<2
order by 2

Total de clientes por países

select country, count(customer_id)
from country join city using (country_id)
join address using (city_id)
join customer using (address_id)
group by country