namespace keyword and __NAMESPACE__ constant

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

PHP supports two ways of abstractly accessing elements within the current namespace, the __NAMESPACE__ magic constant, and the namespace keyword.

The value of __NAMESPACE__ is a string that contains the current namespace name. In global, un-namespaced code, it contains an empty string.

Example #1 __NAMESPACE__ example, namespaced code

<?php
namespace MyProject;

echo
'"', __NAMESPACE__, '"'; // outputs "MyProject"
?>

Example #2 __NAMESPACE__ example, global code

<?php

echo '"', __NAMESPACE__, '"'; // outputs ""
?>
The __NAMESPACE__ constant is useful for dynamically constructing names, for instance:

Example #3 using __NAMESPACE__ for dynamic name construction

<?php
namespace MyProject;

function
get($classname)
{
$a = __NAMESPACE__ . '\\' . $classname;
return new
$a;
}
?>

The namespace keyword can be used to explicitly request an element from the current namespace or a sub-namespace. It is the namespace equivalent of the self operator for classes.

Example #4 the namespace operator, inside a namespace

<?php
namespace MyProject;

use
blah\blah as mine; // see "Using namespaces: Aliasing/Importing"

blah\mine(); // calls function MyProject\blah\mine()
namespace\blah\mine(); // calls function MyProject\blah\mine()

namespace\func(); // calls function MyProject\func()
namespace\sub\func(); // calls function MyProject\sub\func()
namespace\cname::method(); // calls static method "method" of class MyProject\cname
$a = new namespace\sub\cname(); // instantiates object of class MyProject\sub\cname
$b = namespace\CONSTANT; // assigns value of constant MyProject\CONSTANT to $b
?>

Example #5 the namespace operator, in global code

<?php

namespace\func(); // calls function func()
namespace\sub\func(); // calls function sub\func()
namespace\cname::method(); // calls static method "method" of class cname
$a = new namespace\sub\cname(); // instantiates object of class sub\cname
$b = namespace\CONSTANT; // assigns value of constant CONSTANT to $b
?>