Instantiating class by string using PHP 5.3 namespaces

Question

I can't get around an issue instantiating a new class by using a string variable and PHP 5.3. namespaces. For example, this works;

$class = 'Reflection';
$object = new $class();

However, this does not;

$class = '\Application\Log\MyClass';
$object = new $class();

A fatal error gets thrown stating the class cannot be found. However it obviously can be instantiated if using the FQN i.e.;

$object = new \Application\Log\MyClass;

I've found this to be aparrent on PHP 5.3.2-1 but not not in later versions. Is there a work around for this?

Artefacto · Accepted Answer · 2011-02-21 23:23:04Z

71

$class = 'Application\Log\MyClass';
$object = new $class();

The starting \ introduces a (fully qualified) namespaced identifier, but it's not part of the class name itself.

edited Feb 21, 2011 at 23:23

answered Feb 21, 2011 at 23:13

Artefacto

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1 Comment

milkovsky Over a year ago

You can use also $class = __NAMESPACE__ . '\MyClass'; for namespaces

Francesco Casula · Accepted Answer · 2014-03-25 18:59:45Z

5

Another way to achieve the same result but with dynamic arguments is as follows. Please consider the class below as the class you want to instantiate.

<?php

// test.php

namespace Acme\Bundle\MyBundle;

class Test {
    public function __construct($arg1, $arg2) {
        var_dump(
            $arg1,
            $arg2
        );
    }
}

And then:

<?php

require_once('test.php');

(new ReflectionClass('Acme\Bundle\MyBundle\Test'))->newInstanceArgs(['one', 'two']);

If you are not using a recent version of PHP, please use the following code that replaces the last line of the example above:

$r = new ReflectionClass('Acme\Bundle\MyBundle\Test');
$r->newInstanceArgs(array('one', 'two'));

The code will produce the following output:

string(3) "one"
string(3) "two"

edited Mar 25, 2014 at 18:59

answered Mar 25, 2014 at 10:41

Francesco Casula

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Comments

Przemek Kro · Accepted Answer · 2020-03-20 16:41:27Z

0

I had the same problem and I have found some way around this problem. Probably not very efficient, but at least works.

function getController($controller)
{
    return new $controller;
}

$object = getController('Application\Log\MyClass');

answered Mar 20, 2020 at 16:41

Przemek Kro

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