Question

I’m looking for the get_called_class() equivalent for __FILE__ … Maybe something like get_included_file()?

I have a set of classes which would like to know what directory they exist in. Something like this:

<?php

class A {

    protected $baseDir;

    public function __construct() {
        $this->baseDir = dirname(__FILE__);
    }

    public function getBaseDir() {
        return $this->baseDir;
    }
}

?>

And in some other file, in some other folder…

<?php

class B extends A {
    // ...
}

class C extends B {
    // ...
}

$a = new A;
echo $a->getBaseDir();

$b = new B;
echo $b->getBaseDir();

$c = new C;
echo $c->getBaseDir();

// Annnd... all three return the same base directory.

?>

Now, I could do something ghetto, like adding $this->baseDir = dirname(__FILE__) to each and every extending class, but that seems a bit… ghetto. After all, we’re talking about PHP 5.3, right? Isn’t this supposed to be the future?

Is there another way to get the path to the file where a class was declared?

Answer 1

what if you don’t use __FILE__ but a separate variable and set the variable to __FILE__ in each class

class A {

    protected static $baseDir;
    protected $filename = __FILE__; // put this in every file

    public function __construct() {

    }

    public function getBaseDir() {
        return dirname($this->filename) . '<br>'; // use $filename instead of __FILE__
    }   

}

require('bdir/b.php');
require('cdir/c.php');

class B extends A {
    protected $filename = __FILE__; // put this in every file
}

$a = new A;
echo $a->getBaseDir();

$b = new B;
echo $b->getBaseDir();

$c = new C;
echo $c->getBaseDir();

you still have to redeclare the property in each class, but not the method