How can I mention keyword arguments in php constructor? - php

Question

I am new to PHP OOP.
I have this constructor
protected $path='default';
function __construct($var1, $var2 ..){
$this->var1=$var1;
$this->var2=$var2;
}
$obj = MyClass($var1='www')
I want that if I make object and don't pass any arguments then I get the object with empty values.
But if mention the properties in the constructor like above then the object should have those properties set.
Currently if I define the constructor with args and don't supply any then I get error
.

Answer 1

You could make use func_get_args() in PHP
<?php
class foo
{
private $arrParams;
function __construct()
{
if (func_num_args() != 0) {
$this->arrParams = $this->setValues(func_get_args());
}
}
public function setValues($params)
{
return $params;
}
public function dispParams()
{
print_r($this->arrParams);
}
public function retVal($var)
{
return $this->arrParams[$var];
}
}
$foo1 = new Foo(3, 45, 64, 34);
$foo1->dispParams();
OUTPUT :
Array
(
[0] => 3
[1] => 45
[2] => 64
[3] => 34
)
For getting a corresponding value from the array... You could just call
$foo1->retVal(2); // Prints you 64

Answer 2

you need to declare class property like this:
class yourClass{
protected $path='default';
public $var1;
public $var2;
function __construct($var1="dd", $var2='ss'){
$this->var1=$var1;
$this->var2=$var2;
}
}
$obj = new yourClass();
echo "without params: ".$obj->var1.$obj->var2;
$obj2 = new yourClass("sdf","sfdsdf");
echo " with params : ".$obj2->var1.$obj2->var2;
Demo