I want to use an unknown class with a variable in another class:
Example
$classandmethod = "index@show"; namespace Lib\abc; use Lib\xyz; class abc{ function controller($classandmethod) { $a = explode("@", $classandmethod); use Lib\Controller\.$a[0]; } } But maybe it's not true, please help everyone!
From the manual:
The use keyword must be declared in the outermost scope of a file (the global scope) or inside namespace declarations
So you can't use a class where you're trying to do it. Then, to instantiate a class from a different namespace from a variable, rather than useing it, supply the whole namespaced class:
<?php namespace G4\Lib { class A { public $a = "test"; } } namespace G4 { class B { public $a; public function __construct($class) { $class = "\\G4\\Lib\\".$class; $this->a = new $class; // here's the magic } } $b = new B("A"); var_dump($b->a); // test } The main problem is that use cannot have a variable as part of it as it's used when parsing the file and the variable is only available at runtime.
This is an example of how you can do what you seem to be after...
<?php namespace Controller; ini_set('display_errors', 'On'); error_reporting(E_ALL); class Index { public function show() { echo "Showing..."; } } $classandmethod = "Controller\Index@show"; list($className,$method) = explode("@", $classandmethod); $a= new $className(); $a->$method(); This displays...
Showing... You could of course say that all of these classes must be in the Controller namespace and so the code will change to
$classandmethod = "Index@show"; list($className,$method) = explode("@", $classandmethod); $className = "Controller\\".$className; $a= new $className(); $a->$method(); 