PHP 5.6.16 is available

Using namespaces: Aliasing/Importing

The ability to refer to an external fully qualified name with an alias, or importing, is an important feature of namespaces. This is similar to the ability of unix-based filesystems to create symbolic links to a file or to a directory.

PHP namespaces support two kinds of aliasing or importing: aliasing a class name, and aliasing a namespace name. Note that importing a function or constant is not supported.

In PHP, aliasing is accomplished with the use operator. Here is an example showing all 3 kinds of importing:

Example #1 importing/aliasing with the use operator

<?php
namespace foo;
use
My\Full\Classname as Another;

// this is the same as use My\Full\NSname as NSname
use My\Full\NSname;

// importing a global class
use \ArrayObject;

$obj = new namespace\Another// instantiates object of class foo\Another$obj = new Another// instantiates object of class My\Full\Classname
NSname\subns\func(); // calls function My\Full\NSname\subns\func
$a = new ArrayObject(array(1)); // instantiates object of class ArrayObject // without the "use \ArrayObject" we would instantiate an object of class foo\ArrayObject ?> Note that for namespaced names (fully qualified namespace names containing namespace separator, such as Foo\Bar as opposed to global names that do not, such as FooBar), the leading backslash is unnecessary and not allowed, as import names must be fully qualified, and are not processed relative to the current namespace. PHP additionally supports a convenience shortcut to place multiple use statements on the same line Example #2 importing/aliasing with the use operator, multiple use statements combined <?php use My\Full\Classname as AnotherMy\Full\NSname;$obj = new Another// instantiates object of class My\Full\Classname
NSname\subns\func(); // calls function My\Full\NSname\subns\func
?>

Importing is performed at compile-time, and so does not affect dynamic class, function or constant names.

Example #3 Importing and dynamic names

<?php
use My\Full\Classname as AnotherMy\Full\NSname;

$obj = new Another// instantiates object of class My\Full\Classname$a 'Another';
$obj = new$a;      // instantiates object of class Another
?>

In addition, importing only affects unqualified and qualified names. Fully qualified names are absolute, and unaffected by imports.

Example #4 Importing and fully qualified names

<?php
use My\Full\Classname as AnotherMy\Full\NSname;

$obj = new Another// instantiates object of class My\Full\Classname$obj = new \Another// instantiates object of class Another
$obj = new Another\thing// instantiates object of class My\Full\Classname\thing$obj = new \Another\thing// instantiates object of class Another\thing
?>

User Contributed Notes 13 notes

38
k at webnfo dot com
2 years ago
Note that you can not alias global namespace:

use \ as test;

echo test\strlen('');

won't work.
18
x at d dot a dot r dot k dot REMOVEDOTSANDTHIS dot gray dot org
2 years ago
You are allowed to "use" the same resource multiple times as long as it is imported under a different alias at each invocation.

For example:

<?php
use Lend;
use
Lend\l1;
use
Lend\l1 as l3;
use
Lend\l2;
use
Lend\l1\Keller;
use
Lend\l1\Keller as Stellar;
use
Lend\l1\Keller as Zellar;
use
Lend\l2\Keller as Dellar;

...

?>

In the above example, "Keller", "Stellar", and "Zellar" are all references to "\Lend\l1\Keller", as are "Lend\l1\Keller", "l1\Keller", and "l3\Keller".
sernuzh at gmail dot com
6 months ago
You'll get here the
Fatal error: Cannot declare class others\name because the name is already in use
So you can't get two classes <name> inside one namespace
<?php
namespace my {
class
name {
public function
__construct(){
echo
'my_namespace_object';
}
}
}
namespace
others{
use
my\name;
class
name {
public function
__construct(){
echo
'others_namespace_object';
}
}
$newObject = new name(); } ?> 16 c dot 1 at smithies dot org 4 years ago If you are testing your code at the CLI, note that namespace aliases do not work! (Before I go on, all the backslashes in this example are changed to percent signs because I cannot get sensible results to display in the posting preview otherwise. Please mentally translate all percent signs henceforth as backslashes.) Suppose you have a class you want to test in myclass.php: <?php namespace my%space; class myclass { // ... } ?> and you then go into the CLI to test it. You would like to think that this would work, as you type it line by line: require 'myclass.php'; use my%space%myclass; // should set 'myclass' as alias for 'my%space%myclass'$x = new myclass; // FATAL ERROR

I believe that this is because aliases are only resolved at compile time, whereas the CLI simply evaluates statements; so use statements are ineffective in the CLI.

If you put your test code into test.php:
<?php
require 'myclass.php';
use
my%space%myclass;
$x = new myclass; //... ?> it will work fine. I hope this reduces the number of prematurely bald people. 11 anon 1 year ago The <?php use ?> statement does not load the class file. You have to do this with the <?php require ?> statement or by using an autoload function. 10 cl 2 years ago Something that is not immediately obvious, particular with PHP 5.3, is that namespace resolutions within an import are not resolved recursively. i.e.: if you alias an import and then use that alias in another import then this latter import will not be fully resolved with the former import. For example: use \Controllers as C; use C\First; use C\Last; Both the First and Last namespaces are NOT resolved as \Controllers\First or \Controllers\Last as one might intend. kelerest123 at gmail dot com 8 months ago For the fifth example (example #5): When in block scope, it is not an illegal use of use keyword, because it is used for sharing things with traits. Anonymous 2 years ago The last example on this page shows a possibly incorrect attempt of aliasing, but it is totally correct to import a trait \Languages\Languages\Danish. thinice at gmail.com 4 years ago Because imports happen at compile time, there's no polymorphism potential by embedding the use keyword in a conditonal. e.g.: <?php if ($objType == 'canine') {
use
Animal\Canine as Beast;
}
if (
$objType == 'bovine') { use Animal\Bovine as Beast; }$oBeast = new Beast;
$oBeast->feed(); ?> samuel dot roze at gmail dot com 3 years ago (All the backslashes in namespaces are slashes because I can't figure out how to post backslashes here.) You can have the same "use" for a class and a namespace. For example, if you have these files: <?php // foo/bar.php namespace foo; class bar { public function __toString () { return 'foo\bar\__toString()'; } } ?> <?php // foo/bar/MyClass.php namespace foo/bar; class MyClass { public function __toString () { return 'foo\bar\MyClass\__toString()'; } } ?> In another namespace, you can do: <?php namespace another; require_once 'foo/bar.php'; require_once 'foo/bar/MyClass.php'; use foo/bar;$bar = new bar();
echo
$bar."\n";$class = new bar/MyClass();
echo
$class."\n"; ?> And it will makes the following output: foo\bar\__toString() foo\bar\MyClass\__toString() -4 nsdhami at live dot jp 5 years ago The "use" keyword can not be declared inside the function or method. It should be declared as global, after the "namespace" as: <?php namespace mydir; // works perfectly use mydir/subdir/Class1 as Class1; function fun1() { // Parse error: syntax error, unexpected T_USE use mydir/subdir/Class1 as Class1; } class Class2 { public function fun2() { // Parse error: syntax error, unexpected T_USE use mydir/subdir/Class1 as Class1; } } ?> -13 Jan Tvrdk 4 years ago Importing and aliasing an interface name is also supported. -15 Dr. Gianluigi &#34;Zane&#34; Zanettini 9 months ago I was attempting to use something like this: <?php use$my_variable_namespace
?>

This is not supported. I did this instead:

<?php
if(..)
use
My\First\Namespace;
else
use
My\Other\Namespace;
?>