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FAQ: things you need to know about namespaces

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

This FAQ is split into two sections: common questions, and some specifics of implementation that are helpful to understand fully.

First, the common questions.

  1. If I don't use namespaces, should I care about any of this?
  2. How do I use internal or global classes in a namespace?
  3. How do I use namespaces classes functions, or constants in their own namespace?
  4. How does a name like \my\name or \name resolve?
  5. How does a name like my\name resolve?
  6. How does an unqualified class name like name resolve?
  7. How does an unqualified function name or unqualified constant name like name resolve?

There are a few implementation details of the namespace implementations that are helpful to understand.

  1. Import names must not conflict with classes defined in the same file.
  2. Nested namespaces are not allowed.
  3. Dynamic namespace names (quoted identifiers) should escape backslash.
  4. Undefined Constants referenced using any backslash die with fatal error
  5. Cannot override special constants null, true or false

If I don't use namespaces, should I care about any of this?

No. Namespaces do not affect any existing code in any way, or any as-yet-to-be-written code that does not contain namespaces. You can write this code if you wish:

Приклад #1 Accessing global classes outside a namespace

<?php
$a
= new \stdClass;
?>

This is functionally equivalent to:

Приклад #2 Accessing global classes outside a namespace

<?php
$a
= new stdClass;
?>

How do I use internal or global classes in a namespace?

Приклад #3 Accessing internal classes in namespaces

<?php
namespace foo;
$a = new \stdClass;

function
test(\ArrayObject $parameter_type_example = null) {}

$a = \DirectoryIterator::CURRENT_AS_FILEINFO;

// extending an internal or global class
class MyException extends \Exception {}
?>

How do I use namespaces classes, functions, or constants in their own namespace?

Приклад #4 Accessing internal classes, functions or constants in namespaces

<?php
namespace foo;

class
MyClass {}

// using a class from the current namespace as a parameter type
function test(MyClass $parameter_type_example = null) {}
// another way to use a class from the current namespace as a parameter type
function test(\foo\MyClass $parameter_type_example = null) {}

// extending a class from the current namespace
class Extended extends MyClass {}

// accessing a global function
$a = \globalfunc();

// accessing a global constant
$b = \INI_ALL;
?>

How does a name like \my\name or \name resolve?

Names that begin with a \ always resolve to what they look like, so \my\name is in fact my\name, and \Exception is Exception.

Приклад #5 Fully Qualified names

<?php
namespace foo;
$a = new \my\name(); // instantiates "my\name" class
echo \strlen('hi'); // calls function "strlen"
$a = \INI_ALL; // $a is set to the value of constant "INI_ALL"
?>

How does a name like my\name resolve?

Names that contain a backslash but do not begin with a backslash like my\name can be resolved in 2 different ways.

If there is an import statement that aliases another name to my, then the import alias is applied to the my in my\name.

Otherwise, the current namespace name is prepended to my\name.

Приклад #6 Qualified names

<?php
namespace foo;
use
blah\blah as foo;

$a = new my\name(); // instantiates "foo\my\name" class
foo\bar::name(); // calls static method "name" in class "blah\blah\bar"
my\bar(); // calls function "foo\my\bar"
$a = my\BAR; // sets $a to the value of constant "foo\my\BAR"
?>

How does an unqualified class name like name resolve?

Class names that do not contain a backslash like name can be resolved in 2 different ways.

If there is an import statement that aliases another name to name, then the import alias is applied.

Otherwise, the current namespace name is prepended to name.

Приклад #7 Unqualified class names

<?php
namespace foo;
use
blah\blah as foo;

$a = new name(); // instantiates "foo\name" class
foo::name(); // calls static method "name" in class "blah\blah"
?>

How does an unqualified function name or unqualified constant name like name resolve?

Function or constant names that do not contain a backslash like name can be resolved in 2 different ways.

First, the current namespace name is prepended to name.

Finally, if the constant or function name does not exist in the current namespace, a global constant or function name is used if it exists.

Приклад #8 Unqualified function or constant names

<?php
namespace foo;
use
blah\blah as foo;

const
FOO = 1;

function
my() {}
function
foo() {}
function
sort(&$a)
{
\sort($a); // calls the global function "sort"
$a = array_flip($a);
return
$a;
}

my(); // calls "foo\my"
$a = strlen('hi'); // calls global function "strlen" because "foo\strlen" does not exist
$arr = array(1,3,2);
$b = sort($arr); // calls function "foo\sort"
$c = foo(); // calls function "foo\foo" - import is not applied

$a = FOO; // sets $a to value of constant "foo\FOO" - import is not applied
$b = INI_ALL; // sets $b to value of global constant "INI_ALL"
?>

Import names must not conflict with classes defined in the same file.

The following script combinations are legal:

file1.php

<?php
namespace my\stuff;
class
MyClass {}
?>

another.php

<?php
namespace another;
class
thing {}
?>

file2.php

<?php
namespace my\stuff;
include
'file1.php';
include
'another.php';

use
another\thing as MyClass;
$a = new MyClass; // instantiates class "thing" from namespace another
?>

There is no name conflict, even though the class MyClass exists within the my\stuff namespace, because the MyClass definition is in a separate file. However, the next example causes a fatal error on name conflict because MyClass is defined in the same file as the use statement.

<?php
namespace my\stuff;
use
another\thing as MyClass;
class
MyClass {} // fatal error: MyClass conflicts with import statement
$a = new MyClass;
?>

Nested namespaces are not allowed.

PHP does not allow nesting namespaces

<?php
namespace my\stuff {
namespace
nested {
class
foo {}
}
}
?>
However, it is easy to simulate nested namespaces like so:
<?php
namespace my\stuff\nested {
class
foo {}
}
?>

Dynamic namespace names (quoted identifiers) should escape backslash

It is very important to realize that because the backslash is used as an escape character within strings, it should always be doubled when used inside a string. Otherwise there is a risk of unintended consequences:

Приклад #9 Dangers of using namespaced names inside a double-quoted string

<?php
$a
= "dangerous\name"; // \n is a newline inside double quoted strings!
$obj = new $a;

$a = 'not\at\all\dangerous'; // no problems here.
$obj = new $a;
?>
Inside a single-quoted string, the backslash escape sequence is much safer to use, but it is still recommended practice to escape backslashes in all strings as a best practice.

Undefined Constants referenced using any backslash die with fatal error

Any undefined constant that is unqualified like FOO will produce a notice explaining that PHP assumed FOO was the value of the constant. Any constant, qualified or fully qualified, that contains a backslash will produce a fatal error if not found.

Приклад #10 Undefined constants

<?php
namespace bar;
$a = FOO; // produces notice - undefined constants "FOO" assumed "FOO";
$a = \FOO; // fatal error, undefined namespace constant FOO
$a = Bar\FOO; // fatal error, undefined namespace constant bar\Bar\FOO
$a = \Bar\FOO; // fatal error, undefined namespace constant Bar\FOO
?>

Cannot override special constants null, true or false

Any attempt to define a namespaced constant that is a special, built-in constant results in a fatal error

Приклад #11 Undefined constants

<?php
namespace bar;
const
NULL = 0; // fatal error;
const true = 'stupid'; // also fatal error;
// etc.
?>

add a note

User Contributed Notes 6 notes

up
15
manolachef at gmail dot com
12 years ago
There is a way to define a namespaced constant that is a special, built-in constant, using define function and setting the third parameter case_insensitive to false:

<?php
namespace foo;
define(__NAMESPACE__ . '\NULL', 10); // defines the constant NULL in the current namespace
var_dump(NULL); // will show 10
var_dump(null); // will show NULL
?>

No need to specify the namespace in your call to define(), like it happens usually
<?php
namespace foo;
define(INI_ALL, 'bar'); // produces notice - Constant INI_ALL already defined. But:

define(__NAMESPACE__ . '\INI_ALL', 'bar'); // defines the constant INI_ALL in the current namespace
var_dump(INI_ALL); // will show string(3)"bar". Nothing unespected so far. But:

define('NULL', 10); // defines the constant NULL in the current namespace...
var_dump(NULL); // will show 10
var_dump(null); // will show NULL
?>

If the parameter case_insensitive is set to true
<?php
namespace foo;
define (__NAMESPACE__ . '\NULL', 10, true); // produces notice - Constant null already defined
?>
up
7
shaun at slickdesign dot com dot au
8 years ago
When creating classes or calling static methods from within namespaces using variables, you need to keep in mind that they require the full namespace in order for the appropriate class to be used; you CANNOT use an alias or short name, even if it is called within the same namespace. Neglecting to take this into account can cause your code to use the wrong class, throw a fatal missing class exception, or throw errors or warnings.

In these cases, you can use the magic constant __NAMESPACE__, or specify the full namespace and class name directly. The function class_exists also requires the full namespace and class name, and can be used to ensure that a fatal error won't be thrown due to missing classes.

<?php

namespace Foo;
class
Bar {
public static function
test() {
return
get_called_class();
}
}

namespace
Foo\Foo;
class
Bar extends \Foo\Bar {
}

var_dump( Bar::test() ); // string(11) "Foo\Foo\Bar"

$bar = 'Foo\Bar';
var_dump( $bar::test() ); // string(7) "Foo\Bar"

$bar = __NAMESPACE__ . '\Bar';
var_dump( $bar::test() ); // string(11) "Foo\Foo\Bar"

$bar = 'Bar';
var_dump( $bar::test() ); // FATAL ERROR: Class 'Bar' not found or Incorrect class \Bar used
up
3
theking2 at king dot ma
2 years ago
Just like class names currently namespaces are not case sensitive. So no errors will be shown here:

<?php declare(strict_types=1);
namespace
Foo;
class
Bar {
public function
__construct() {
echo
'Map constructed';
}
}

$foobar = new \foo\bar();
up
3
teohad at NOSPAM dot gmail dot com
8 years ago
[Editor's note: that behavior is caused by a bug in PHP 7.0, which has been fixed as of PHP 7.0.7.]

Regarding the entry "Import names cannot conflict with classes defined in the same file".
- I found that since PHP 7.0 this is no longer the case.
In PHP 7.0 you can have a class with a name that matches an imported class (or namespace or both at the same time).

<?php
namespace ns1 {
class
ns1 {
public static function
write() {
echo
"ns1\\ns1::write()\n";
}
}
}

namespace
ns1\ns1 {
class
ns1c {
public static function
write() {
echo
"ns1\\ns1\\ns1c::write()\n";
}
}
}

namespace
ns2 {
use
ns1\ns1 as ns1; // both a class in ns1, and a namespace ns1\ns1

// the next class causes fatal error in php 5.6, not in 7.0
class ns1 {
public static function
write() {
echo
"ns2\\ns1::write()\n";
}
}

ns1::write(); // calls imported ns1\ns1::write()
ns1\ns1c::write(); // calls imported ns1\ns1\ns1c::write()
namespace\ns1::write(); // calls ns2\ns1::write()
}
?>
up
7
phpcoder
9 years ago
Regarding "Neither functions nor constants can be imported via the use statement." Actually you can do it in PHP 5.6+:

<?php

// importing a function (PHP 5.6+)
use function My\Full\functionName;

// aliasing a function (PHP 5.6+)
use function My\Full\functionName as func;

// importing a constant (PHP 5.6+)
use const My\Full\CONSTANT;
?>
up
-5
okaresz
11 years ago
To correct manolachef's answer: define() ALWAYS defines constants in the GLOBAL namespace.

As nl-x at bita dot nl states in the note at http://www.php.net/manual/en/function.define.php, the constant "NULL" can be defined with define() case-sensitively, but can only be retrieved with constant(), leaving the meaning of NULL uppercase keyword as the only value of the type null.
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