PHP: Namespaces e recursos de linguagem dinâmica

Namespaces e recursos de linguagem dinâmica

(PHP 5 >= 5.3.0, PHP 7, PHP 8)

A implementação de namespaces do PHP é influenciada por sua natureza dinâmica como linguagem de programação. Assim, para converter código como o exemplo a seguir em código com namespace:

Exemplo #1 Acessando elementos dinamicamente

exemplo1.php:

<?php
class nomedaclasse
{
    function __construct()
    {
        echo __METHOD__,"\n";
    }
}
function nomedafuncao()
{
    echo __FUNCTION__,"\n";
}
const nomedaconstante = "global";

$a = 'nomedaclasse';
$obj = new $a; / imprime nomedaclasse::__construct
$b = 'nomedafuncao';
$b(); / imprime nomedafuncao
echo constant('nomedaconstante'), "\n"; / imprime global
?>

Deve-se usar o nome totalmente qualificado (nome da classe prefixado com o namespace). Observe que, como não há diferença entre um nome qualificado e um nome totalmente qualificado em um nome de classe, nome de função ou nome de constante dinâmico, a barra invertida inicial não é necessária.

Exemplo #2 Acessando dinamicamente elementos com namespace

<?php
namespace nomedonamespace;
class nomedaclasse
{
    function __construct()
    {
        echo __METHOD__,"\n";
    }
}
function nomedafuncao()
{
    echo __FUNCTION__,"\n";
}
const nomedaconstante = "comnamespace";

/* note que, ao usar aspas duplas, "\\nomedonamespace\\nomedaclasse" deve ser usado */
$a = '\nomedonamespace\nomedaclasse';
$obj = new $a; / imprime nomedonamespace\nomedaclasse::__construct
$a = 'nomedonamespace\nomedaclasse';
$obj = new $a; / também imprime nomedonamespace\nomedaclasse::__construct
$b = 'nomedonamespace\nomedafuncao';
$b(); / imprime nomedonamespace\nomedafuncao
$b = '\nomedonamespace\nomedafuncao';
$b(); / também imprime nomedonamespace\nomedafuncao
echo constant('\nomedonamespace\nomedaconstante'), "\n"; / imprime comnamespace
echo constant('nomedonamespace\nomedaconstante'), "\n"; / também imprime comnamespace
?>

Certifique-se de ler a nota sobre escapamento de namespaces em strings.

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Notas Enviadas por Usuários (em inglês) 8 notes

down

Alexander Kirk ¶

13 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)

down

Daan ¶

5 years 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

down

museyib dot e at gmail dot com ¶

6 years 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.
?>

down

guilhermeblanco at php dot net ¶

15 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



?>

down

akhoondi+php at gmail dot com ¶

11 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 ...
?>

down

m dot mannes at gmail dot com ¶

8 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!";
?>

down

anisgazig at gmail dot com ¶

3 years ago

<?php

/single or double quotes with single or double backslash in dynamic namespace class.

namespace Country_Name{
class Mexico{
function __construct(){
echo __METHOD__,"<br>";
}

$a = 'Country_Name\Mexico';/Country_Name\Mexico::__construct
$a = "Country_Name\Mexico";
/Country_Name\Mexico::__construct
$a = '\Country_Name\Mexico';
/Country_Name\Mexico::__construct
$a = "\Country_Name\Mexico";
/Country_Name\Mexico::__construct
$a = "\\Country_Name\\Mexico";
/Country_Name\Mexico::__construct
$o = new $a;

}

/* if your namespace name or class name start with lowercase n then you should be alart about the use of single or double quotes with backslash  */

namespace name_of_country{
class Japan{
function __construct()
{
    echo __METHOD__,"<br>";
}

}

$a = 'name_of_country\Japan';
/name_of_country\Japan::__construct
$a = "name_of_country\Japan";
/name_of_country\Japan::__construct
$a = '\name_of_country\Japan';
/name_of_country\Japan::__construct
/$a = "\name_of_country\Japan";
/Fatal error: Uncaught Error: Class ' ame_of_country\Japan' not found 
/In this statement "\name_of_country\Japan" means -first letter n with "\ == new line("\n). for fix it we can use double back slash or single quotes with single backslash. 
$a = "\\name_of_country\\Japan";
/name_of_country\Japan::__construct
$o = new $a;
}

namespace Country_Name{
class name{
function __construct(){
echo __METHOD__,"<br>";
}

$a = 'Country_Name\name';
/Country_Name\Norway::__construct
$a = "Country_Name\name";
/Country_Name\Norway::__construct
$a = '\Country_Name\name';
/Country_Name\Norway::__construct
/$a = "\Country_Name\name";
/Fatal error: Uncaught Error: Class '\Country_Name ame' not found 

/In this statement "\Country_Name\name" at class name's first letter n with "\ == new line("\n). for fix it we can use double back slash or single quotes with single backslash
$a = "\\Country_Name\\name";
/Country_Name\name::__construct
$o = new $a;

}

/"\n == new line are case insensitive so "\N could not affected

 ?>

down

scott at intothewild dot ca ¶

15 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

?>

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