PHP Instance object with path

Question

I'm trying to create an instance of an object with a path like(C:\wamp\www...)

In a new project I have this method, and I try with that to instance an object of an another project.

public function getControllerObject($class)
{
    $object =  null;
    $class = realpath($class);  
    $class = str_replace('.php','',$class);

    $object = new $class();

}

the variable $class have for exemple this value :

C:\wamp\www\myproject\projectBundle\Controller\DefaultController

But i get FatalErrorException: Error: Class. Class not found

I already try to put 2 backslash but it doesn't work.

Any idea?

Are you trying to create an auto loader or something, because that's what it sounds like. — xlordt
– xlordt, Commented Oct 22, 2013 at 13:40
Yes i'm trying to get all the controller class of a bundle symfony2. Maybe there is others solution for do this. — altore
– altore, Commented Oct 22, 2013 at 13:43
Symfony2 brings its own autoloader. No need to create a new one. Annotate them with a use statement and just create them like $controller = new IndexController(); — ferdynator
– ferdynator, Commented Oct 22, 2013 at 14:34
You have an error because you try to create an object from the path C:\wamp\[...]\Controller\DefaultController . The PHP file must be included or called by the autoloader if you want to access it. — A.L
– A.L, Commented Oct 22, 2013 at 14:54
consider adding a require_once(realpath($class)); statement to your code, then add $object = new ClassName();. You can't create an instance of a file, you can, however, create an instance of a class definition — Elias Van Ootegem
– Elias Van Ootegem, Commented Oct 23, 2013 at 11:23

Nico · Accepted Answer · 2013-10-22 13:37:45Z

1

Use basename, not realpath:

$class = basename($class);

For your example it should produce DefaultController.

answered Oct 22, 2013 at 13:37

Nico

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3 Comments

altore Over a year ago

Ok, but the class is not on the same namespace

Nico Over a year ago

So what is your namespace, the project name? You could use a combination of basename and dirname to build the dynamic namespace and class string.

altore Over a year ago

the namespace for the class is myproject\projectBundle\Controller. But my code for instance this object is on a separate project

dbrumann · Accepted Answer · 2013-10-22 14:53:01Z

First of all, what you are trying to do sounds like a horrible idea! If you want to access the class from another project you should add it (the project or the files) to your dependencies.

Your problem is, that what you basically do is:

$object = new C:\wamp\www\myproject\projectBundle\Controller\DefaultController();

which is obviously not what you want. When you do not enter a valid path (which I assume is what happened) realpath will return false and it won't work either.

You have to somehow determine which part of the path is part of the namespace and which is not. You could do this by also adding the project root, meaning which part to strip from the path or the namespace itself, i.e. the part to keep from the path (assuming you follow the recommendations from PSR-0), which should leave you with what you want (after str_replace() a preg_replace() or something similar).

Anyway, the cleanest way to solve your problem (apart from adding a dependency), is to use Symfony Class Loader instead of hardcoding paths into your application.

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