(PHP 5 >= 5.3.0, PHP 7, PHP 8)
允许通过别名引用或导入外部的完全限定名称，是命名空间的一个重要特征。这有点类似于在类 unix 文件系统中可以创建对其它的文件或目录的符号连接。
所有支持命名空间的 PHP 版本支持三种别名或导入方式：为类名称使用别名、为接口使用别名或为命名空间名称使用别名。PHP 5.6 开始允许导入函数或常量或者为它们设置别名。
在PHP中，别名是通过操作符 use 来实现的. 下面是一个使用所有可能的五种导入方式的例子：
Example #1 使用 use 操作符导入/使用别名
<?php
namespace foo;
use My\Full\Classname as Another;
// 下面的例子与 use My\Full\NSname as NSname 相同
use My\Full\NSname;
// 导入一个全局类
use ArrayObject;
// 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;
$obj = new namespace\Another; // 实例化 foo\Another 对象
$obj = new Another; // 实例化 My\Full\Classname 对象
NSname\subns\func(); // 调用函数 My\Full\NSname\subns\func
$a = new ArrayObject(array(1)); // 实例化 ArrayObject 对象
// 如果不使用 "use \ArrayObject" ，则实例化一个 foo\ArrayObject 对象
func(); // calls function My\Full\functionName
echo CONSTANT; // echoes the value of My\Full\CONSTANT
?>
Foo\Bar 以及相对的不包含命名空间分隔符的全局名称如
FooBar）来说，前导的反斜杠是不必要的也不推荐的，因为导入的名称必须是完全限定的，不会根据当前的命名空间作相对解析。
为了简化操作，PHP 还支持在一行中使用多个 use 语句
Example #2 通过 use 操作符导入/使用别名，一行中包含多个 use 语句
<?php
use My\Full\Classname as Another, My\Full\NSname;
$obj = new Another; // 实例化 My\Full\Classname 对象
NSname\subns\func(); // 调用函数 My\Full\NSname\subns\func
?>
导入操作是在编译执行的，但动态的类名称、函数名称或常量名称则不是。
Example #3 导入和动态名称
<?php
use My\Full\Classname as Another, My\Full\NSname;
$obj = new Another; // 实例化一个 My\Full\Classname 对象
$a = 'Another';
$obj = new $a; // 实际化一个 Another 对象
?>
另外，导入操作只影响非限定名称和限定名称。完全限定名称由于是确定的，故不受导入的影响。
Example #4 导入和完全限定名称
<?php
use My\Full\Classname as Another, My\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
?>
The use keyword must be declared in the
outermost scope of a file (the global scope) or inside namespace
declarations. This is because the importing is done at compile
time and not runtime, so it cannot be block scoped. The following
example will show an illegal use of the use
keyword:
Example #5 Illegal importing rule
<?php
namespace Languages;
class Greenlandic
{
use Languages\Danish;
...
}
?>
Note:
Importing rules are per file basis, meaning included files will NOT inherit the parent file's importing rules.
Bear in mind that it's perfectly fine to alias namespaces, ie:
<?php
use A\B\C\D\E\User;
new User();
?>
can be also written as:
<?php
use A\B\C\D\E as ENamespace;
new ENamespace\User();
?>
however following will not work:
<?php
use A\B\C\D\E as ENamespace;
use ENamespace\User;
new User();
?>
> PHP Error: Class "ENamespace\User" not found
Note: you can import not existed items without errors:
<?php
use UndefinedClass;
use function undefined_fn;
use const UNDEFINED_CONST;
?>
but you cant use/call they:
<?php
$new UndefinedClass; // Error: Use of undefined class
use function undefined_fn; // Error: Use of undefined function
use const UNDEFINED_CONST; // Error: Use of undefined constant
?>
namespace test{
use test\xyz\tikku;
use test\xyz\tikku as class_alias;
use function test\xyz\tikku\foo;
use function test\xyz\tikku\foo as func_alias;
use const test\xyz\tikku\ABC;
use const test\xyz\tikku\ABC as const_alias;
$obj=new tikku;
$obj->Display(); // I am in test\xyz namespace
$obj=new tikku\dhairya;
$obj->Display(); // I am in test\xyz\tikku namespace
$obj=new class_alias\dhairya;
$obj->Display(); // I am in test\xyz\tikku namespace
$obj=new \class_alias\dhairya;
$obj->Display(); // I am in class_alias namespace
}
namespace test\xyz{
class tikku{
function Display(){
echo "I am in ".__namespace__." namespace<br/><hr/>";
}
}
}
namespace test\xyz\tikku{
class dhairya{
function Display(){
echo "I am in ".__namespace__." namespace<br/><hr/>";
}
}
}
namespace class_alias{
class dhairya{
function Display(){
echo "I am in ".__namespace__." namespace<br/><hr/>";
}
}
}
amazing!!
use function strval as numberToString;
var_dump(numberToString(123));
//print
//string(3) "123"
For those hoping for an easy fix of DRY with {} unfortunately you can't nest it or do anything cool with it. In case you can only have one per use and if you don't have one then the whole use is assumed to be wrapped in brackets. That means if you do have one you can't use , as normal with use!
Not possible:
use A\B\C\{
D, E,
F\{X, Y, Z}
},
X\Y\Z,
H\{
I\{
Y\Y\Y\Y\Y,
Z, H, E
},
J\{
A\{
G\H\J
B\N
},
G\H\J
}
};
Here is a handy way of importing classes, functions and conts using a single use keyword:
<?php
use Mizo\Web\ {
Php\WebSite,
Php\KeyWord,
Php\UnicodePrint,
JS\JavaScript,
function JS\printTotal,
function JS\printList,
const JS\BUAIKUM,
const JS\MAUTAM
};
?>
I couldn't find answer to this question so I tested myself.
I think it's worth noting:
<?php
use ExistingNamespace\NonExsistingClass;
use ExistingNamespace\NonExsistingClass as whatever;
use NonExistingNamespace\NonExsistingClass;
use NonExistingNamespace\NonExsistingClass as whatever;
?>
None of above will actually cause errors unless you actually try to use class you tried to import.
<?php
// And this code will issue standard PHP error for non existing class.
use ExistingNamespace\NonExsistingClass as whatever;
$whatever = new whatever();
?>
In Chinese,there is an error in translation:
// 如果不使用 "use \ArrayObject" ，则实例化一个 foo\ArrayObject 对象
it should be
// 如果不使用 "use ArrayObject" ，则实例化一个 foo\ArrayObject 对象
/*********************************************/
中文下翻译有错误
// 如果不使用 "use \ArrayObject" ，则实例化一个 foo\ArrayObject 对象
这句话应该是
// 如果不使用 "use ArrayObject" ，则实例化一个 foo\ArrayObject 对象
Note that "use" importing/aliasing only applies to the current namespace block.
<?php
namespace SuperCoolLibrary
{
class Meta
{
static public function getVersion()
{
return '2.7.1';
}
}
}
namespace
{
use SuperCoolLibrary\Meta;
echo Meta::getVersion();//outputs 2.7.1
}
namespace
{
echo Meta::getVersion();//fatal error
}
?>
To get the expected behavior, you'd use:
class_alias('SuperCoolLibrary\Meta','Meta');
To clarify the distinction between inserting a trait in a class and importing a trait in a namespace, here is an example where we first import and then insert a trait.
<?php
namespace ns1;
trait T {
static $a = "In T";
}
namespace ns2;
use ns1\T; // Importing the name of trait ns1\T in the namespace ns2
class C {
use T; // Inserting trait T in the class C, making use of the imported name.
}
namespace main;
use ns2\C;
echo C::$a; // In T;
The keyword "use" has been recycled for three distinct applications:
1- to import/alias classes, traits, constants, etc. in namespaces,
2- to insert traits in classes,
3- to inherit variables in closures.
This page is only about the first application: importing/aliasing. Traits can be inserted in classes, but this is different from importing a trait in a namespace, which cannot be done in a block scope, as pointed out in example 5. This can be confusing, especially since all searches for the keyword "use" are directed to the documentation here on importing/aliasing.
Note the code `use ns1\c1` may refer to importing class `c1` from namespace `ns1` as well as importing whole namespace `ns1\c1` or even import both of them in one line. Example:
<?php
namespace ns1;
class c1{}
namespace ns1\c1;
class c11{}
namespace main;
use ns1\c1;
$c1 = new c1();
$c11 = new c1\c11();
var_dump($c1); // object(ns1\c1)#1 (0) { }
var_dump($c11); // object(ns1\c1\c11)#2 (0) { }
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.
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.
Note that you can not alias global namespace:
use \ as test;
echo test\strlen('');
won't work.
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.
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.
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".
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;
$x = new myclass;
//...
?>
it will work fine.
I hope this reduces the number of prematurely bald people.
Because imports happen at compile time, there's no polymorphism potential by embedding the use keyword in a conditonal.
e.g.:
<?php
if ($objType == 'canine') {
use Animal\Canine as Beast;
}
if ($objType == 'bovine') {
use Animal\Bovine as Beast;
}
$oBeast = new Beast;
$oBeast->feed();
?>