Attempted to load class "Class" from the global namespace. Error Symfony2

Question

Hi I have a class name "Customer" in a variable like:

$myclass = "Customer";

Now I have created object for this class at run time inside a service file:

namespace MyBundle\Service;

use Doctrine\ORM\EntityManager;
use Doctrine\ORM\QueryBuilder;
use MyBundle\Component\Data\handle\Customer;
use Symfony\Component\HttpFoundation\Request;

class MyServices
{

    private $em;   

    public function __construct(EntityManager $entityManager)
    {
        $this->em = $entityManager;        
}

public function getClassCustomer($className)
{

   $object = new $className();
    }
}

Now I am getting following error:

Attempted to load class "Customer" from the global namespace

Even Customer class is already defined and included in the same file:

Please suggest what could be the Issue. thanks in advance

Youre explanation is a bit chaotic. Please post minimal but full code snippet that will let us reproduce the issue. use/BundleName/ClassFolder: is not including anything. — Jakub Matczak
– Jakub Matczak, Commented Oct 3, 2016 at 10:19
Hi dragoste: actualy namespace Mybundle\Service; use Mybundle\Component\DataFolder\Classess\Customer; I am using on the top of file: — Sunil Rawat
– Sunil Rawat, Commented Oct 3, 2016 at 10:23

Jakub Matczak · Accepted Answer · 2016-10-03 10:50:13Z

1

The issue is that when you're using variable as class name, then use statement doesn't apply.

When you do

use MyBundle\Component\Data\handle\Customer;
new Customer();

It's resolved as

new MyBundle\Component\Data\handle\Customer();

But with this:

use MyBundle\Component\Data\handle\Customer;
$className = "Customer";        
$object = new $className();

It's still just:

$object = new \Customer();

Take a look Example #3 on this page as it's similar case:

use My\Full\Classname as Another, My\Full\NSname;

$obj = new Another; // instantiates object of class My\Full\Classname
$a = 'Another';
$obj = new $a;      // instantiates object of class Another

answered Oct 3, 2016 at 10:50

Jakub Matczak

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2 Comments

Sunil Rawat Over a year ago

My class name is dynamic: it will be any class like customer and some time user etc.. then in this case?

Jakub Matczak Over a year ago

You need full qualified class name in your variable (with full namespace).

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