Namespaces und dynamische Sprachfeatures

(PHP 5 >= 5.3.0, PHP 7)

Die Implementierung von Namespaces in PHP ist stark von seinen Eigenschaften als Programmiersprache mit dynamischen Features beeinflusst. Man kann also den folgenden Code in Code mit Namespaces umformen:

Beispiel #1 Dynamischer Zugriff auf Elemente

example1.php:

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

$a 'classname';
$obj = new $a// gibt classname::__construct aus
$b 'funcname';
$b(); // gibt funcname aus
echo constant('constname'), "\n"// gibt global aus
?>
Um diesen Code umzuformen, muss man den vollständig qualifizierten Namen (Klassenname mit namespace-Präfix) verwenden. Beachten Sie, dass, weil es keinen Unterschied zwischen einem qualifizierten und vollständig qualifizierten Namen innerhalb eines dynamischen Klassen-, Funktions- oder Konstantennamen gibt, der führende Backslash nicht notwendigerweise angegeben werden muss.

Beispiel #2 Dynamischer Zugriff auf Elemente mit Namespace

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

/* Wenn man doppelte Anführungszeichen verwendet,
muss "\\namespacename\\classname" verwendet werden */
$a '\namespacename\classname';
$obj = new $a// gibt namespacename\classname::__construct aus
$a 'namespacename\classname';
$obj = new $a// gibt ebenfalls namespacename\classname::__construct aus
$b 'namespacename\funcname';
$b(); // gibt namespacename\funcname aus
$b '\namespacename\funcname';
$b(); // gibt ebenfalls namespacename\funcname aus
echo constant('\namespacename\constname'), "\n"// gibt namespaced aus
echo constant('namespacename\constname'), "\n"// gibt ebenfalls namespaced aus
?>

Bitte lesen Sie auch den Hinweis zum Escaping von Namespacenamen in Strings.

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User Contributed Notes 4 notes

up
45
Alexander Kirk
231 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
11
guilhermeblanco at php dot net
233 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
5
akhoondi+php at gmail dot com
229 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
3
scott at intothewild dot ca
233 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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