jQuery style Constructors in PHP - php

Question

Is there a way to instantiate a new PHP object in a similar manner to those in jQuery? I'm talking about assigning a variable number of arguments when creating the object. For example, I know I could do something like:
...
//in my Class
__contruct($name, $height, $eye_colour, $car, $password) {
...
}
$p1 = new person("bob", "5'9", "Blue", "toyota", "password");
But I'd like to set only some of them maybe. So something like:
$p1 = new person({
name: "bob",
eyes: "blue"});
Which is more along the lines of how it is done in jQuery and other frameworks. Is this built in to PHP? Is there a way to do it? Or a reason I should avoid it?

Answer 1

the best method to do this is using an array:
class Sample
{
private $first = "default";
private $second = "default";
private $third = "default";
function __construct($params = array())
{
foreach($params as $key => $value)
{
if(isset($this->$key))
{
$this->$key = $value; //Update
}
}
}
}
And then construct with an array
$data = array(
'first' => "hello"
//Etc
);
$Object = new Sample($data);

Answer 2

class foo {
function __construct($args) {
foreach($args as $k => $v) $this->$k = $v;
echo $this->name;
}
}
new foo(array(
'name' => 'John'
));
The closest I could think of.
If you want to be more fancy and just want to allow certain keys, you can use __set() (only on php 5)
var $allowedKeys = array('name', 'age', 'hobby');
public function __set($k, $v) {
if(in_array($k, $this->allowedKeys)) {
$this->$k = $v;
}
}

Answer 3

get args won't work as PHP will see only one argument being passed.
public __contruct($options) {
$options = json_decode( $options );
....
// list of properties with ternary operator to set default values if not in $options
....
}
have a looksee at json_decode()

Answer 4

The closest I can think of is to use array() and extract().
...
//in your Class
__contruct($options = array()) {
// default values
$password = 'password';
$name = 'Untitled 1';
$eyes = '#353433';
// extract the options
extract ($options);
// stuff
...
}
And when creating it.
$p1 = new person(array(
'name' => "bob",
'eyes' => "blue"
));