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:

Exemplo #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.

Exemplo #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.

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User Contributed Notes 7 notes

up
60
Alexander Kirk
8 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)
up
17
guilhermeblanco at php dot net
10 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

?>
up
2
m dot mannes at gmail dot com
2 years ago
Case you are trying call a static method that's the way to go:

<?php
class myClass
{
    public static function
myMethod()
    {
      return
"You did it!\n";
    }
}

$foo = "myClass";
$bar = "myMethod";

echo
$foo::$bar(); // prints "You did it!";
?>
up
4
akhoondi+php at gmail dot com
6 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 ...
?>
up
2
museyib dot e at gmail dot com
7 months ago
Be careful when using dynamic accessing namespaced elements. If you use double-quote backslashes will be parsed as escape character.

<?php
    $a
="\namespacename\classname"; //Invalid use and Fatal error.
   
$a="\\namespacename\\classname"; //Valid use.
   
$a='\namespacename\classname'; //Valid use.
?>
up
1
scott at intothewild dot ca
10 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

?>
up
1
Daan
26 days ago
Important to know is that you need to use the *fully qualified name* in a dynamic class name. Here is an example that emphasizes the difference between a dynamic class name and a normal class name.

<?php
namespace namespacename\foo;

class
classname     
{                                                                                       
    function
__construct()                                                              
    {                                                                                   
        echo
'bar';
    }                                                                                   
}                                                                                       

$a = '\namespacename\foo\classname'; // Works, is fully qualified name                  
$b = 'namespacename\foo\classname'; // Works, is treated as it was with a prefixed "\"  
$c = 'foo\classname'; // Will not work, it should be the fully qualified name           

// Use dynamic class name                                                                                        
new $a; // bar
new $b; // bar
new $c; // [500]: / - Uncaught Error: Class 'foo\classname' not found in

// Use normal class name                                                                                        
new \namespacename\foo\classname; // bar
new namespacename\foo\classname; // [500]: / - Uncaught Error: Class 'namespacename\foo\namespacename\foo\classname' not found
new foo\classname; // [500]: / - Uncaught Error: Class 'namespacename\foo\foo\classname' not found
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