Unable to access variable from included file - php

Question

As the title says there is a problem accessing variable (associative array) inside class from included file. Here is the source code both class and include file:
require("applications/cw_database.php");
require("config/dbConfig.php");
require("config/appConfig.php");
class APP_ASSESMENTS
{
private $dbObj;
private $DisplayOutput = "";
public function __construct($PageParams)
{
try
{
$dbObj = new CW_DB($dbConfig['hostname'],$dbConfig['username'],$dbConfig['password'],$dbConfig['name'],$dbConfig['port']);
} catch (Exception $e) {
throw new ErrorException($e);
}
}
...
The other part has nothing to do with $dbConfig.
Also this is the included file (config/dbConfig.php):
/*
Testing configuration for MySQL database
*/
$dbConfig['username'] = "phpcoursework"; // changed on demand
$dbConfig['password'] = "phpcoursework"; // changed on demand
$dbConfig['hostname'] = "localhost"; // changed on demand
$dbConfig['name'] = "students"; // changed on demand
$dbConfig['port'] = 3306; // default for MySQL

Answer 1

First, $dbObj will not automatically assume class member scope, it will create a local copy of CW_DB and discard it when __construct returns. You need to explicitly reference the property;
$this->dbObj = ...
Anyway, global state using global as suggested by others will "work", but if you're using OOP practices you're best not to do that. You can actually return from an include(), so an option would be to do the following:
// your config file dbConfig.php
return array(
'username' => "phpcoursework",
'password' => "phpcoursework",
'hostname' => "localhost",
'name' => "students",
'port' => 3306,
);
And inject it into the object, via constructor or method (here's constructor)
class APP_ASSESMENTS
{
private $dbObj;
public function __construct($dbConfig, $PageParams)
{
$dbObj = new CW_DB($dbConfig['hostname'], $dbConfig['username'],
$dbConfig['password'], $dbConfig['name'], $dbConfig['port']);
// ...
}
}
// include() here, will actually return the array from the config file
$appAssesments = new \APP_ASSESMENTS(include('dbConfig.php'), $PageParams);
It would be recommended that you go another level up: instead, inject the database object itself, taking the dependency out of your APP_ASSESSMENTS class.
(Also, PascalCase is the typical convention of class naming, such as AppAssessments and CwDb)
$dbObj = new CwDb(include('dbConfig.php'));
$appAssessments = new AppAssessments($dbObj, $etc, $etc);
This simple change allows you to remove the dependency from AppAssessments on CwDb. That way, if you extend CwDb for some reason, you can just pass in an instance of the extended class without having to change any code in AppAssessments
You can change the AppAssessments constructor like so:
public function __construct(CwDb $db, $etc, $etc){
$this->db = $db;
// ...
}
This takes advantage of PHPs (limited, albeit still useful) type-hinting, ensuring the first argument is always of the correct type.
This plays into part of the open/closed principle: classes should be open to extension but closed for modification.

Answer 2

Includes are used in the scope of access. So, to access the variables you need to include the files within your class. As earlier mentioned global will let you access variables from another scope too. But global should be used with caution! See the documentation.
See the manual for more information.
Edit: I need to make it clear that globals are never a good alternative for handling these kind of critical variables..

Answer 3

public function __construct($PageParams){
global $dbConfig;
try{
$dbObj = new CW_DB($dbConfig['hostname'],$dbConfig['username'],$dbConfig['password'],$dbConfig['name'],$dbC onfig['port']);
} catch (Exception $e) {
throw new ErrorException($e);
}
}
or you could use $GLOBALS['dbConfig'].