Class not found when successfully included - php

Question

I get this fatal error for class not being found:
Fatal error: Uncaught Error: Class 'HomeController' not found
But the file is included.
require_once self::CONTROLLERS_PATH . $this->controller . '.php';
var_dump(file_exists(self::CONTROLLERS_PATH . $this->controller . '.php'));
new $this->controller;
I debugged it by var_dumping file_exists which returns true. I checked the namespaces and everything looks good. If I var dump $this->controller, its value is as intended - string(14) "HomeController". If I hard code it like new HomeController();, the class is initialized. Why is it not working from the variable, though? You can see my code below. This is my file structure:
App.php
<?php
namespace App\Core;
class App
{
protected $controller = 'HomeController';
protected $method = 'index';
protected $params = [];
const CONTROLLERS_PATH = '../App/Controllers/';
public function __construct ()
{
$url = $this->parseUrl();
if(file_exists(self::CONTROLLERS_PATH . ucfirst($url[0]) . '.php'))
{
$this->controller = ucfirst($url[0]); //Update $this->controller value
unset($url[0]);
echo 'file exists';
}
require_once self::CONTROLLERS_PATH . $this->controller . '.php';
var_dump(file_exists(self::CONTROLLERS_PATH . $this->controller . '.php'));
var_dump($this->controller);
new $this->controller;
if(isset($url[1]))
{
if(method_exists($this->controller, $url[1]))
{
$this->method = $url[1];
unset($url[1]);
echo 'method exists';
}
}
$this->params = $url ? array_values($url): [];
call_user_func_array([$this->controller, $this->method], $this->params);
var_dump($url);
}
public function parseUrl()
{
if(isset($_GET['url']))
{
return $url = explode('/', filter_var(rtrim($_GET['url'], '/'), FILTER_SANITIZE_URL));
}
}
}
HomeController.php
<?php
namespace App\Controllers;
class HomeController
{
public function index()
{
echo 'asd';
}
}

Answer 1

When instantiating a class via a variable name you always have to include the whole namespace. And this has to be stored as a string first.
An example:
namespace Api;
Class test {
public function __construct() {
echo "test";
}
}
$className = 'test';
$classNameNamespaced = "\\Api\\".$className;
$a = new $classNameNamespaced; // echoes "test"
in your case:
$className = "\\App\\Controllers\\".$this->controller;
new $className;

Answer 2

Because the code is in a namespace, new $this->controller is different than new HomeController even if the value of $this->controller is HomeController.
The expression new HomeController is evaluated at the compile time and, because of the namespace, it is expanded to new \App\Controllers\HomeController (rule #6 on the Name resolution rules). The interpreter needs the fully qualified class name in order to find it.
The expression new $this->controller is evaluated during runtime and it leads to new HomeController which is the same as new \HomeController. There isn't any class HomeController defined in the root namespace, hence the error you get.
You can make it easily work by building the full class name into a string:
$class = 'App\\Controllers\\'.$this->controller;
new $class;
Find more examples in the documentation page "Namespaces and dynamic language features".
But the best way to load classes is to use the autoloader generated by Composer.