PHP - How can I check a class has no arguments constructor? - php

Question

In few worlds : I would like to do the same but in PHP.
In details : I have a Class A method that instantiates a Class X which can be a Class B or C.
Class A
class A{
...
protected function init(){
if( !empty( $this->sub_pages ) ){
foreach ( $this->sub_pages as $sub_page ){
$class = $sub_page['class_path'];
//Here, I need to check if ClassX ( = $class) constructor has arguments.
if( no arguments){
new $class();
}else{
new $class( $sub_page['data'] );
}
}
}
}
...
}
Class B
class B{
public function __construct(){ //<-- No arguments
}
}
Class C
class C extends D{
public function __construct( $data ){ //<-- With arguments
parent::__construct( $data );
}
}
Someone know the answer ?

Answer 1

If you want to check whether a class has a constructor and if that constructor accepts any params or not then you can do it using PHP's Reflection Class for example:
$reflector = new \ReflectionClass('SomeClass');
$constructor = $reflector->getConstructor();
if ($constructor && $constructor->getParameters()) {
// Since your class needs $sub_page['data'] and
// you already have this in your current scope
$instance = $reflector->newInstanceArgs($sub_page['data']);
} else {
$instance = new SomeClass;
}
Btw, If you have type hinted dependencies (like other class instance) then you can find out what is the dependency and can also new up that dependent class to pass as param.

Answer 2

Base on #The Alpha answer, I found this :
class A{
...
protected function init(){
if( !empty( $this->sub_pages ) ){
foreach ( $this->sub_pages as $sub_page ){
try {
$reflector = new \ReflectionClass( $class );
if (!$constructor = $reflector->getConstructor()) {
printf( "The Class '%s' has not got a constructor", $class );
}else {
// has a constructor
if ( $paramsArray = $constructor->getParameters() ) {
new $class( $sub_page['data'] );
}else{
new $class();
}
}
} catch ( \ReflectionException $e ) {
echo $e;
}
}
}
...
}