命名空间和动态语言特征

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

PHP 命名空间的实现受到其语言自身的动态特征的影响。因此，如果要将下面的代码转换到命名空间中：

Example #1 动态访问元素

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
?>
必须使用完全限定名称（包括命名空间前缀的类名称）。注意因为在动态的类名称、函数名称或常量名称中，限定名称和完全限定名称没有区别，因此其前导的反斜杠是不必要的。

Example #2 动态访问命名空间的元素

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

include 'example1.php';

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

/* 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
?>

请一定别忘了阅读 对字符串中的命名空间名称转义的注解.

User Contributed Notes

Daan 19-Sep-2019 08:56
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
museyib dot e at gmail dot com 09-Mar-2019 04:48
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.
?>
m dot mannes at gmail dot com 25-Feb-2017 06:36
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!";
?>
akhoondi+php at gmail dot com 09-Aug-2013 04:17
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 ...
?>
Alexander Kirk 06-Jul-2011 12:57
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)
scott at intothewild dot ca 07-Aug-2009 03:33
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

?>
guilhermeblanco at php dot net 16-Jun-2009 12:04
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

?>
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