Fatal error while extending a class which does not have namespace - php

Question

I have two classes, Database and States. The class Database has no namespace assigned. The class States, written in file states.php, has a namespace assigned as below:
namespace Resources;
require_once 'database.php';
$states = new States();
echo $states->readAll();
class States extends Database{
private $db;
public function __construct(){
$this->db = new Database();
}
function readAll(){
return $this->db->Execute('SELECT * FROM states;');
}
}
If I enter the url http://localhost/mydomain/objects/states.php which should instantiate the class and show list of states, an error is thrown as:
Fatal error: Uncaught Error: Class 'Resources\Database' not found in
C:\xampp\htdocs\mydomain\objects\states.php on line 5
If I remove the line namespace Resources; the code runs.
I have recently decided to implemen namespaces in my website, so most likely, I am doing something wrong.
Any help please?

Answer 1

The PHP manual has a section introducing namespaces which is well worth reading. In particular, have a look at this page drawing the analogy to a file system:
The same principle can be applied to namespaced elements in PHP. For
example, a class name can be referred to in three ways:
Unqualified name, or an unprefixed class name like $a = new foo(); or foo::staticmethod();. If the current namespace is currentnamespace,
this resolves to currentnamespace\foo. If the code is global,
non-namespaced code, this resolves to foo.
Qualified name, or a prefixed class name like $a = new subnamespace\foo(); or subnamespace\foo::staticmethod();. If the
current namespace is currentnamespace, this resolves to
currentnamespace\subnamespace\foo. If the code is global,
non-namespaced code, this resolves to subnamespace\foo.
Fully qualified name, or a prefixed name with global prefix operator like $a = new \currentnamespace\foo(); or
\currentnamespace\foo::staticmethod();. This always resolves to the
literal name specified in the code, currentnamespace\foo.
In your example, Database is an "unqualified name", so point 1 applies. The current namespace is Resources, so it refers to Resources\Database. If you remove the namespace Resources; line, the code becomes "global, non-namespaced code", so it refers to just Database.
To refer to the class Database when the current namespace is Resources, you can use a "fully qualified name", as described in point 3. So in this case, you would write class States extends \Database.
There are other ways to write it, such as adding use statements to import/alias a name, but there's no point me copying the whole PHP manual here.