How can I retrieve a class namespace automatically?
The magic var __NAMESPACE__ is unreliable since in subclasses it's not correctly defined.
Example:
class Foo\bar\A -> __NAMESPACE__ === Foo\bar
class Ping\pong\B extends Foo\bar\A -> __NAMESPACE__ === Foo\bar (it should be 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?
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 ;)
The namespace of class Foo\Bar\A is Foo\Bar, so the __NAMESPACE__ is working very well. What you are looking for is probably namespaced classname that you could easily get by joining echo __NAMESPACE__ . '\\' . __CLASS__;.
Consider next example:
namespace Foo\Bar\FooBar;
use Ping\Pong\HongKong;
class A extends HongKong\B {
function __construct() {
echo __NAMESPACE__;
}
}
new A;
Will print out Foo\Bar\FooBar which is very correct...
And even if you then do
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...
EDIT: If you need to get the namespace of the nested class within the main that is nesting it, simply use:
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), '\\'));
}
}
In PHP 5.5, ::class is available which makes things 10X easier. E.g.
A::class
Use Reflection class.
$class_name = get_class($this);
$reflection_class = new \ReflectionClass($class_name);
$namespace = $reflection_class->getNamespaceName();
Related
So I created these two classes
//Quarter.php
namespace Resources;
class Quarter {
...
}
//Epoch.php
namespace Resources;
class Epoch {
public static function initFromType($value, $type) {
$class = "Quarter";
return new $class($value, $type);
}
}
Now this is a a very simplified version of both, but is enough to illustrate my question. The classes as they are shown here will not work as it will not find the Quarter class. To make it work I could change the $class variable to
$class = "\Resources\Quarter";
So my question is: Why do I need to use the namespace here when both classes are already members of the same namespace. The namespace is only needed when I put the classname in a variable so doing:
public static function initFromType($value, $type) {
return new Quarter($value, $type);
}
will work without problems. Why is this and is there any potential traps here I need to avoid?
Because strings can be passed around from one namespace to another. That makes name resolution ambiguous at best and easily introduces weird problems.
namespace Foo;
$class = 'Baz';
namespace Bar;
new $class; // what class will be instantiated?
A literal in a certain namespace does not have this problem:
namespace Foo;
new Baz; // can't be moved, it's unequivocally \Foo\Baz
Therefore, all "string class names" are always absolute and need to be written as FQN:
$class = 'Foo\Baz';
(Note: no leading \.)
You can use this as shorthand, sort of equivalent to a self-referential self in classes:
$class = __NAMESPACE__ . '\Baz';
Is it possible to declare a class and have it extend a variable?
class Child extends $parentClass {}
Yes, it is with eval. But it is not recommended.
<?php
function dynamic_class_name() {
if(time() % 60)
return "Class_A";
if(time() % 60 == 0)
return "Class_B";
}
eval(
"class MyRealClass extends " . dynamic_class_name() . " {" .
# some code string here, possibly read from a file
. "}"
);
?>
Is Eval an evil?! Read this.
no you cant.
this is because in a normal definition:
class A extends SomeOtherClass {
}
SomeOtherClass is a namespace reference, not an instance of the class. And you cant use a variable to provide that namespace because variables in PHP are defined at runtime, whereas classes are defined in the compile phase.
But... if youre using PHP7, you can do this:
$newClass = new class extends SomeOtherClass {};
which is not exactly what you want, but goes some way.
A Class can not extend from a variable , You need to wrap the variable inside a class to extend . Hope this helps
I know this is an old question, but I hope my answer helps someone.
You can use class_alias() to an "intermediate" class:
class_alias($parentClass, 'ParentClassAlias');
class Child extends ParentClassAlias {}
So I created these two classes
//Quarter.php
namespace Resources;
class Quarter {
...
}
//Epoch.php
namespace Resources;
class Epoch {
public static function initFromType($value, $type) {
$class = "Quarter";
return new $class($value, $type);
}
}
Now this is a a very simplified version of both, but is enough to illustrate my question. The classes as they are shown here will not work as it will not find the Quarter class. To make it work I could change the $class variable to
$class = "\Resources\Quarter";
So my question is: Why do I need to use the namespace here when both classes are already members of the same namespace. The namespace is only needed when I put the classname in a variable so doing:
public static function initFromType($value, $type) {
return new Quarter($value, $type);
}
will work without problems. Why is this and is there any potential traps here I need to avoid?
Because strings can be passed around from one namespace to another. That makes name resolution ambiguous at best and easily introduces weird problems.
namespace Foo;
$class = 'Baz';
namespace Bar;
new $class; // what class will be instantiated?
A literal in a certain namespace does not have this problem:
namespace Foo;
new Baz; // can't be moved, it's unequivocally \Foo\Baz
Therefore, all "string class names" are always absolute and need to be written as FQN:
$class = 'Foo\Baz';
(Note: no leading \.)
You can use this as shorthand, sort of equivalent to a self-referential self in classes:
$class = __NAMESPACE__ . '\Baz';
So I created these two classes
//Quarter.php
namespace Resources;
class Quarter {
...
}
//Epoch.php
namespace Resources;
class Epoch {
public static function initFromType($value, $type) {
$class = "Quarter";
return new $class($value, $type);
}
}
Now this is a a very simplified version of both, but is enough to illustrate my question. The classes as they are shown here will not work as it will not find the Quarter class. To make it work I could change the $class variable to
$class = "\Resources\Quarter";
So my question is: Why do I need to use the namespace here when both classes are already members of the same namespace. The namespace is only needed when I put the classname in a variable so doing:
public static function initFromType($value, $type) {
return new Quarter($value, $type);
}
will work without problems. Why is this and is there any potential traps here I need to avoid?
Because strings can be passed around from one namespace to another. That makes name resolution ambiguous at best and easily introduces weird problems.
namespace Foo;
$class = 'Baz';
namespace Bar;
new $class; // what class will be instantiated?
A literal in a certain namespace does not have this problem:
namespace Foo;
new Baz; // can't be moved, it's unequivocally \Foo\Baz
Therefore, all "string class names" are always absolute and need to be written as FQN:
$class = 'Foo\Baz';
(Note: no leading \.)
You can use this as shorthand, sort of equivalent to a self-referential self in classes:
$class = __NAMESPACE__ . '\Baz';
I posted some questions previously regarding the use of Namespaces in PHP and from what I got, this example code I have below should be working.
However I am getting errors when I try to use Namespace in PHP like this. Here is the first error when running the code below as is...
Fatal error: Class 'Controller' not found in E:\Controllers\testing.php on line 6
E:\Controller\testing.php File
<?php
use \Controller;
include('testcontroller.php');
$controller = new Controller;
$controller->show();
?>
E:\Controller\testcontroller.php File
<?php
use \Library\Registry;
namespace Controller
{
class Controller
{
public $registry;
function __construct()
{
include('E:\Library\Registry.class.php');
$this->registry = new Registry;
}
function show()
{
echo $this->registry;
echo '<br>Registry was ran inside testcontroller.php<br>';
}
}
}
?>
E:\Library\Registry.class.php File
<?php
namespace Library\Registry
{
class Registry
{
function __construct()
{
return 'Registry.class.php Constructor was ran';
}
}
}
?>
As you can see I tried to make it as simple as possible just to get the Namespace part working. I have tried different variations and cannot seem to figure it out.
Even when using use statement, you need to specify the namespace of the class you are trying to instantiate. There are a lot of examples here: http://www.php.net/manual/en/language.namespaces.importing.php
To understand it better, I will describe to you how it works. In your case, when you do use \Controller, the whole Controller namespace becomes available to you, but not the classes that are in this namespace. So, for example:
<?php
include('testcontroller.php');
use \Controller;
// Desired class is in namespace!
$controller = new Controller\Controller();
// Error, because in current scope there is no such class
$controller = new Controller();
$controller->show();
?>
Another example:
testcontoller.php:
<?php
namespace Some\Path\To\Controller;
class Controller
{
function __construct()
{
}
function show()
{
echo '<br>Was run inside testcontroller.php<br>';
}
}
?>
testing.php:
<?php
include('testcontroller.php');
use \Some\Path\To\Controller;
// We now can access Controller using only Controller namespace,
// not Some\Path\To\Controller
$controller = new Controller\Controller();
// Error, because, again, in current scope there is no such class
$controller = new Controller();
$controller->show();
?>
If you wish to import exactly the Controller class, you need to do use Controller\Controller - then this class will be accessible in your current scope.
Its not that good idea to name the namespace, like the class, because it is confusing (and I think this is what happens here). There moment you define the alias via use Controller this referenes to either a class \Controller, or the namespace \Controller, but your class, because it is within the namespace, is named \Controller\Controller 1
use Controller;
$class = new Controller\Controller;
or
$class = new \Controller\Controller;
or
use Controller\Controller;
$class = new Controller;
The idea is, that the moment you try to access a class with its relative name it tries to map the "first part" against any alias defined using use (remeber use MyClass is the same as use MyClass as MyClass. The thing after as is the alias).
namespace MyNamespace\MyPackage\SomeComponent\And\So\On {
class MyClass {}
}
namespace Another {
use MyNamespace\MyPackage\SomeComponent; // as SomeComponent
$class = new SomeComponent\An\So\On\MyClass;
}
As you can see PHP finds SomeComponent as the first part and maps it against the SomeComponent-alias the line above.
You can read more about it in the manual about namespaces.
1 Its called "Full-qualified classname", if you name a class with its complete name.
When you put a class Controller in the namespace Controller, then you have to reference it that way:
$controller = new Controller\Controller();
\Controller would be a class in the global (default) namespace, i.e. as if you used no namespace at all.
Strangely I have found that in my example code from the Question above, if I change all the Namespace's that are defined to something like MyLibrary so it would be like this code below...
E:\Library\Registry.class.php File
<?php
namespace MyLibrary
{
class Registry
{
function __construct()
{
echo 'Registry.class.php Constructor was ran';
}
}
}
?>
Then when I use use MyLibrary\Registry; in another file, I am able to access it how I had planned...
$this->registry = new Registry;
The reason this is very strange to me is this now makes a class name appear to be a Namespace as well. So I would not need to set a Namespace to 'MyLibrary\Library' to access the Registry instead I would do it like I showed in this answer to be able to access it with just calling the name of the class.
I hope this makes sense and helps someone else. I will not accept this as the answer as I am hoping someone with more know-how will come in and post a better Answer with explanation
try
<?php
use \Library\Registry;
namespace Controller;
class Controller
{
public $registry;
function __construct()
{
include('E:\Library\Registry.class.php');
$this->registry = new Registry;
}
function show()
{
echo $this->registry;
echo '<br>Registry was ran inside testcontroller.php<br>';
}
}
?>
and
<?php
namespace Library\Registry;
class Registry
{
function __construct()
{
return 'Registry.class.php Constructor was ran';
}
}
?>
First off, I believe you are using composer or composer is initialised in your project. If so, check composer.json file for your autoload, psr-4 definition. For example, if the root of your application is "App", then in your psr-4, you should be doing "autoload": { "psr-4": { "App\\": "./" } },
Furthermore, remember to clear composer cache and dump-autoload from the terminal as follows:
composer clear-cache
composer dump-autoload