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.
Alexander Kirk (2011-07-06 00:57:08)
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();
$c->tell(); // "bar"
?>
(it seems that the namespace-backslashes are stripped from the source code in the preview, maybe it works in the main view. If not: fooA was written as \foo\A and barB as bar\B)
scott at intothewild dot ca (2009-08-07 15:33:34)
as noted by guilhermeblanco at php dot net,
<?php
// fact.php
namespace foo;
class fact {
public function create($class) {
return new $class();
}
}
?>
<?php
// bar.php
namespace foo;
class bar {
...
}
?>
<?php
// index.php
namespace foo;
include('fact.php');
$foofact = new fact();
$bar = $foofact->create('bar'); // attempts to create \bar
// even though foofact and
// bar reside in \foo
?>
guilhermeblanco at php dot net (2009-06-16 12:04:00)
Please be aware of FQCN (Full Qualified Class Name) point.
Many people will have troubles with this:
<?php
// File1.php
namespace foo;
class Bar { ... }
function factory($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
?>