Namespaces and dynamic language features

(PHP 5 >= 5.3.0, PHP 7)

PHP's implementation of namespaces is influenced by its dynamic nature as a programming language. Thus, to convert code like the following example into namespaced code:

Example #1 Dynamically accessing elements

example1.php:


<?php
class classname
{
    function __construct()
    {
        echo __METHOD__,"\n";
    }
}
function funcname()
{
    echo __FUNCTION__,"\n";
}
const constname = "global";

$a = 'classname';
$obj = new $a; // prints classname::__construct
$b = 'funcname';
$b(); // prints funcname
echo constant('constname'), "\n"; // prints global
?>

One must use the fully qualified name (class name with namespace prefix). Note that because there is no difference between a qualified and a fully qualified Name inside a dynamic class name, function name, or constant name, the leading backslash is not necessary.

Example #2 Dynamically accessing namespaced elements


<?php
namespace namespacename;
class classname
{
    function __construct()
    {
        echo __METHOD__,"\n";
    }
}
function funcname()
{
    echo __FUNCTION__,"\n";
}
const constname = "namespaced";

/* note that if using double quotes, "\\namespacename\\classname" must be used */
$a = '\namespacename\classname';
$obj = new $a; // prints namespacename\classname::__construct
$a = 'namespacename\classname';
$obj = new $a; // also prints namespacename\classname::__construct
$b = 'namespacename\funcname';
$b(); // prints namespacename\funcname
$b = '\namespacename\funcname';
$b(); // also prints namespacename\funcname
echo constant('\namespacename\constname'), "\n"; // prints namespaced
echo constant('namespacename\constname'), "\n"; // also prints namespaced
?>

Be sure to read the note about escaping namespace names in strings.

User Contributed Notes

Alexander Kirk

4 years ago


When extending a class from another namespace that should instantiate a class from within the current namespace, you need to pass on the namespace.

<?php // File1.php
namespace foo;
class A {
    public function factory() {
        return new C;
    }
}
class C {
    public function tell() {
        echo "foo";
    }
}
?>

<?php // File2.php
namespace bar;
class B extends \foo\A {}
class C {
    public function tell() {
        echo "bar";
    }
}
?>

<?php
include "File1.php";
include "File2.php";
$b = new bar\B;
$c = $b->factory();
$c->tell(); // "foo" but you want "bar"
?>

You need to do it like this:

When extending a class from another namespace that should instantiate a class from within the current namespace, you need to pass on the namespace.

<?php // File1.php
namespace foo;
class A {
    protected $namespace = __NAMESPACE__;
    public function factory() {
        $c = $this->namespace . '\C';
        return new $c;
    }
}
class C {
    public function tell() {
        echo "foo";
    }
}
?>

<?php // File2.php
namespace bar;
class B extends \foo\A {
    protected $namespace = __NAMESPACE__;
}
class C {
    public function tell() {
        echo "bar";
    }
}
?>

<?php
include "File1.php";
include "File2.php";
$b = new bar\B;
$c = $b->factory();
$c->tell(); // "bar"
?>

(it seems that the namespace-backslashes are stripped from the source code in the preview, maybe it works in the main view. If not: fooA was written as \foo\A and barB as bar\B)

guilhermeblanco at php dot net

6 years ago


Please be aware of FQCN (Full Qualified Class Name) point.

Many people will have troubles with this:



<?php



// File1.php

namespace foo;



class Bar { ... }



function factory($class) {

    return new $class;

}



// File2.php

$bar = \foo\factory('Bar'); // Will try to instantiate \Bar, not \foo\Bar



?>



To fix that, and also incorporate a 2 step namespace resolution, you can check for \ as first char of $class, and if not present, build manually the FQCN:



<?php



// File1.php

namespace foo;



function factory($class) {

    if ($class[0] != '\\') {

        echo '->';

         $class = '\\' . __NAMESPACE__ . '\\' . $class;

    }



    return new $class();

}



// File2.php

$bar = \foo\factory('Bar'); // Will correctly instantiate \foo\Bar



$bar2 = \foo\factory('\anotherfoo\Bar'); // Wil correctly instantiate \anotherfoo\Bar



?>

akhoondi+php at gmail dot com

2 years ago


It might make it more clear if said this way: 

One must note that when using a dynamic class name, function name or constant name, the "current namespace", as in http://www.php.net/manual/en/language.namespaces.basics.php is global namespace.

One situation that dynamic class names are used is in 'factory' pattern. Thus, add the desired namespace of your target class before the variable name.

namespaced.php
<?php
// namespaced.php
namespace Mypackage;
class Foo {
    public function factory($name, $global = FALSE)
    {
        if ($global)
            $class = $name;
        else
            $class = 'Mypackage\\' . $name;
        return new $class;
    }
}

class A {
    function __construct()
    {
        echo __METHOD__ . "<br />\n";
    }
}
class B {
    function __construct()
    {
        echo __METHOD__ . "<br />\n";
    }
}
?>

global.php
<?php 
// global.php
class A {
    function __construct()
    {
        echo  __METHOD__;
    }
}
?>

index.php
<?php
//  index.php
namespace Mypackage;
include('namespaced.php');
include('global.php');
  
  $foo = new Foo();
  
  $a = $foo->factory('A');        // Mypackage\A::__construct 
  $b = $foo->factory('B');        // Mypackage\B::__construct
  
  $a2 = $foo->factory('A',TRUE);    // A::__construct
  $b2 = $foo->factory('B',TRUE);    // Will produce : Fatal error: Class 'B' not found in ...namespaced.php on line ...
?>

scott at intothewild dot ca

6 years ago


as noted by guilhermeblanco at php dot net, 

<?php

  // fact.php

  namespace foo;

  class fact {

    public function create($class) {
      return new $class();
    }
  }

?>

<?php 

  // bar.php

  namespace foo;

  class bar {
  ... 
  }

?>

<?php

  // index.php
 
  namespace foo;

  include('fact.php');
  
  $foofact = new fact();
  $bar = $foofact->create('bar'); // attempts to create \bar
                                  // even though foofact and
                                  // bar reside in \foo

?>