Question

If I have a class (e.g Database) can I check the location of the script that tried to instantiate this class.

An example:

I have an example dir structure like so:

lib
  Database.class.php
app
  action.class.php

and in action.class.php:

$db = new Database;

and in Database.class.php

function __construct(){
$name = //some code to find out that app/action.class.php just tried to instantiate me
$name = explode('/',$name);
if($name[0] = 'app') //allow script to use the class
else $this->__destruct; //dont allow other scripts not in app dir for example
}

I dont know if this is possible or not

Regards
Luke

UPDATE: It seems like this is possible but not good practice (from answers below) thank you for your time, I wont be implimenting this mainly due to @Gordons argument.

Answer 1

If the calling instance or it’s location in the filesystem is a dependency in the Database class, then pass it in through the constructor. If you need find out to where the file is located, use Reflection or the magic __FILE__ constant, e.g.

// action.class.php 
public function initDb()
{
    $this->db = new Database($this);
}

// Database.php
public function __construct($invoker)
{
    // reflective way to get $invokerLocation
    $reflector = new ReflectionClass($invoker);
    $invokerLocation = $reflector->getFileName();
    // check location
}

or simply

// action.class.php 
public function initDb()
{
    $this->db = new Database(__FILE__);
}

// Database.php
public function __construct($invokerLocation)
{
    // check location
}

You do not want debug_backtrace to find out what called the Database constructor. Like the name implies, it is for debugging purposes. Depending on the size of the trace stack, fetching the backtrace can consume a lot of time and resources. Passing it is cleaner and faster.

In addition, if you make security of your code depend on where files are located, then you introduce a dependency on the filesystem, which is code smell to me.