动态获取 PHP 类命名空间
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Get PHP class namespace dynamically
提问by daveoncode
How can I retrieve a class namespace automatically?
如何自动检索类命名空间？
The magic var __NAMESPACE__is unreliable since in subclasses it's not correctly defined.
魔术变量__NAMESPACE__不可靠，因为在子类中它没有正确定义。
Example:
例子：
class Foo\bar\A-> __NAMESPACE__=== Foo\bar
class Foo\bar\A-> __NAMESPACE__=== Foo\bar
class Ping\pong\B extends Foo\bar\A-> __NAMESPACE__=== Foo\bar (it should be Ping\pong)
class Ping\pong\B extends Foo\bar\A-> __NAMESPACE__=== Foo\bar (应该是Ping\pong)
ps: I noticed the same wrong behavior using __CLASS__, but I solved using get_called_class()... is there something like get_called_class_namespace()? How can I implement such function?
ps：我注意到使用了同样的错误行为__CLASS__，但我解决了使用get_called_class()......有没有类似的东西get_called_class_namespace()？我怎样才能实现这样的功能？
UPDATE:
I think the solution is in my own question, since I realized get_called_class()returns the fully qualified class name and thus I can extract the namespace from it :D
...Anyway if there is a more effective approach let me know ;)
更新：
我认为解决方案是在我自己的问题中，因为我意识到get_called_class()返回完全限定的类名，因此我可以从中提取命名空间：D ...无论如何，如果有更有效的方法，请告诉我；）
回答by shadyyx
The namespace of class Foo\Bar\Ais Foo\Bar, so the __NAMESPACE__is working very well. What you are looking for is probably namespaced classnamethat you could easily get by joining echo __NAMESPACE__ . '\\' . __CLASS__;.
class Foo\Bar\Ais的命名空间Foo\Bar，因此__NAMESPACE__运行良好。您正在寻找的可能是命名空间类名，您可以通过join轻松获得echo __NAMESPACE__ . '\\' . __CLASS__;。
Consider next example:
考虑下一个例子：
namespace Foo\Bar\FooBar;
use Ping\Pong\HongKong;
class A extends HongKong\B {
function __construct() {
echo __NAMESPACE__;
}
}
new A;
Will print out Foo\Bar\FooBarwhich is very correct...
将打印出来Foo\Bar\FooBar这是非常正确的...
And even if you then do
即使你那样做
namespace Ping\Pong\HongKong;
use Foo\Bar\FooBar;
class B extends FooBar\A {
function __construct() {
new A;
}
}
it will echo Foo\Bar\FooBar, which again is very correct...
它会回声Foo\Bar\FooBar，这又是非常正确的......
EDIT:If you need to get the namespace of the nested class within the main that is nesting it, simply use:
编辑：如果您需要在嵌套它的 main 中获取嵌套类的命名空间，只需使用：
namespace Ping\Pong\HongKong;
use Foo\Bar\FooBar;
class B extends FooBar\A {
function __construct() {
$a = new A;
echo $a_ns = substr(get_class($a), 0, strrpos(get_class($a), '\'));
}
}
回答by David Lin
In PHP 5.5, ::class is available which makes things 10X easier. E.g.
A::class
在 PHP 5.5 中，::class 可用，这使事情变得简单 10 倍。例如
A::class
