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:

示例 #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; // 输出 classname::__construct
$b = 'funcname';
$b(); // 输出 funcname
echo constant('constname'), "\n"; // 输出 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。

示例 #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";

/* 注意,如果使用双引号,要这样写 "\\namespacename\\classname" */
$a = '\namespacename\classname';
$obj = new $a; // 输出 namespacename\classname::__construct
$a = 'namespacename\classname';
$obj = new $a; // 也会输出 namespacename\classname::__construct
$b = 'namespacename\funcname';
$b(); // 输出 namespacename\funcname
$b = '\namespacename\funcname';
$b(); // 也会输出 namespacename\funcname
echo constant('\namespacename\constname'), "\n"; // 输出 namespaced
echo constant('namespacename\constname'), "\n"; // 也会输出 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.

add a note

User Contributed Notes 8 notes

up
75
Alexander Kirk
12 years ago
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)
up
11
Daan
4 years ago
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
up
6
museyib dot e at gmail dot com
4 years ago
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.
?>
up
16
guilhermeblanco at php dot net
14 years ago
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

?>
up
6
akhoondi+php at gmail dot com
10 years ago
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 ...
?>
up
2
m dot mannes at gmail dot com
7 years ago
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!";
?>
up
2
anisgazig at gmail dot com
2 years ago
<?php

//single or double quotes with single or double backslash in dynamic namespace class.

namespace Country_Name{
class
Mexico{
function
__construct(){
echo
__METHOD__,"<br>";
}
}

$a = 'Country_Name\Mexico';//Country_Name\Mexico::__construct
$a = "Country_Name\Mexico";
//Country_Name\Mexico::__construct
$a = '\Country_Name\Mexico';
//Country_Name\Mexico::__construct
$a = "\Country_Name\Mexico";
//Country_Name\Mexico::__construct
$a = "\\Country_Name\\Mexico";
//Country_Name\Mexico::__construct
$o = new $a;

}

/* if your namespace name or class name start with lowercase n then you should be alart about the use of single or double quotes with backslash */

namespace name_of_country{
class
Japan{
function
__construct()
{
echo
__METHOD__,"<br>";
}

}

$a = 'name_of_country\Japan';
//name_of_country\Japan::__construct
$a = "name_of_country\Japan";
//name_of_country\Japan::__construct
$a = '\name_of_country\Japan';
//name_of_country\Japan::__construct
//$a = "\name_of_country\Japan";
//Fatal error: Uncaught Error: Class ' ame_of_country\Japan' not found
//In this statement "\name_of_country\Japan" means -first letter n with "\ == new line("\n). for fix it we can use double back slash or single quotes with single backslash.
$a = "\\name_of_country\\Japan";
//name_of_country\Japan::__construct
$o = new $a;
}

namespace
Country_Name{
class
name{
function
__construct(){
echo
__METHOD__,"<br>";
}
}

$a = 'Country_Name\name';
//Country_Name\Norway::__construct
$a = "Country_Name\name";
//Country_Name\Norway::__construct
$a = '\Country_Name\name';
//Country_Name\Norway::__construct
//$a = "\Country_Name\name";
//Fatal error: Uncaught Error: Class '\Country_Name ame' not found

//In this statement "\Country_Name\name" at class name's first letter n with "\ == new line("\n). for fix it we can use double back slash or single quotes with single backslash
$a = "\\Country_Name\\name";
//Country_Name\name::__construct
$o = new $a;

}

//"\n == new line are case insensitive so "\N could not affected

?>
up
1
scott at intothewild dot ca
14 years ago
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

?>
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