(PHP 5 >= 5.3.0, PHP 7)
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：
Example #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; // prints classname::__construct
$b = 'funcname';
$b(); // prints funcname
echo constant('constname'), "\n"; // prints global
?>
Example #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";
include 'example1.php';
$a = 'classname';
$obj = new $a; // prints classname::__construct
$b = 'funcname';
$b(); // prints funcname
echo constant('constname'), "\n"; // prints global
/* note that if using double quotes, "\\namespacename\\classname" must be used */
$a = '\namespacename\classname';
$obj = new $a; // prints namespacename\classname::__construct
$a = 'namespacename\classname';
$obj = new $a; // also prints namespacename\classname::__construct
$b = 'namespacename\funcname';
$b(); // prints namespacename\funcname
$b = '\namespacename\funcname';
$b(); // also prints namespacename\funcname
echo constant('\namespacename\constname'), "\n"; // prints namespaced
echo constant('namespacename\constname'), "\n"; // also prints 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.
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.
?>
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!";
?>
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 ...
?>
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();
$c->tell(); // "bar"
?>
(it seems that the namespace-backslashes are stripped from the source code in the preview, maybe it works in the main view. If not: fooA was written as \foo\A and barB as bar\B)
as noted by guilhermeblanco at php dot net,
<?php
// fact.php
namespace foo;
class fact {
public function create($class) {
return new $class();
}
}
?>
<?php
// bar.php
namespace foo;
class bar {
...
}
?>
<?php
// index.php
namespace foo;
include('fact.php');
$foofact = new fact();
$bar = $foofact->create('bar'); // attempts to create \bar
// even though foofact and
// bar reside in \foo
?>
Please be aware of FQCN (Full Qualified Class Name) point.
Many people will have troubles with this:
<?php
// File1.php
namespace foo;
class Bar { ... }
function factory($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
?>