I'm trying to learn the namespaces feature in PHP, however how do I access classes that are in namespaces?
Like, say I have the class Users inside the namespace called Core, how do I access that namespace from the Pages namespace?
I believe this is what you're after:
<?php
$users = new \Core\Users;
echo $users->all();
When you want to use a class that's inside a namespace, you need to define the "absolute path" to the class, like I have done in the example. Note the \ before the Core namespace, that tells PHP to use the Core namespace that is located in the root or "global" namespace of PHP.
So if you wanted to access the Users class in your Pages namespace, you would do the following:
<?php
namespace Pages;
$users = new \Core\Users;
echo $users->all();
There's also another way to use the Users class, which is:
<?php
namespace Pages;
use \Core\Users as Users;
$users = new Users;
echo $users->all();
The use \Core\Users; line allows you to use the Users class from the Core namespace as if it were a normal class inside the Pages namespace.
Related
I created a new FBLogin.php file in App Folder, added class:
namespace App\FBLogin;
class authlogin {}
Now i want to use this class in my Controller File, so i added:
use App\FBLogin\authlogin;
Now when i am trying to use this class authlogin, it is showing me error Class 'App\FBLogin\authlogin' not found
Is there something i am doing wrong?
Laravel Version: 5.5
Why would you use a lowercase format when naming your classes? Anyway, your namespace inside your app folder should follow your file structure.
If you create your class like below,
namespace App\FBLogin;
class authlogin {
// code here
}
Your file structure must be:
app/
FBLogin/
authlogin.php
Then you can use the class anywhere in your app by declaring the proper namespace
use App\FBLogin\authlogin;
$authlogin = new authlogin();
I'm trying to trick PHP into taking a class from another namespace when trying to create a specific class.
I have two class called "page", the first is in the Core namespace:
namespace Core;
class Page {...}
The second inherits from Core\Page, but adds a few things. It is in the Addons namespace.
namespace Addons;
class Page extends \Core\Page{...}
The reason I want to do this is because I want to build my system with an easy addon engine. Whenever I want, I can add a line in an XML file that tells the autoloading function to take the class in the addon namespace instead of the core namespace.
However, when I try to do this :
spl_autoload_register('loadClass');
public function loadClass(string $className)
{
if (Addon_exist_and_is_registered($className))
{
require "/Addons/$className.php";
}
else
{
require "/Core/$className.php";
}
}
$page = new \Core\Page(); <-- error here
I get an error saying that the class \Core\Page cannot be found in the file Addons\Page.php. This is normal behaviour since the class is not in the same namespace and as such, the fully qualified name cannot find the right class.
Is it possible to trick PHP into thinking that a child class in another namespace is actually the right class? I tried this for the addons class;
namespace Core;
class Page extends \Core\Page{...}
But it breaks the inheritance as you cannot inherit yourself.
Ignore that the classes have the "same name". Because they don't. One class is called Core\Page, the other is called Addons\Page. Those are their names, their fully qualified names to be exact. It's as much a difference as Foo and Bar. If you tell PHP to instantiate Core\Page, then it's going to do that; you can't "trick" it into instantiating Addons\Page, since that's an entirely different class name.
Don't try to "trick" anyone, make your system actually extensible and explicitly allow overriding of class names:
$class = 'Core\Page';
if (...) {
$class = 'Addons\Page';
}
$page = new $class;
I would like to Use the same class Helper, located at App\Helpers\Helper, in two different files within the same namespace.
For example:
Class A:
<?php
namespace App\Services;
Use Helper;
class A {...}
Class B:
<?php
namespace App\Services;
Use Helper;
class B {...}
However, this is not allowed. The error is:
Cannot use App\Helpers\Helper as Helper because the name is already in use
I could rename the Helper class in my second file and say Use Helper as SomethingElse, but this seems like a messy solution especially if I want to use this Helper in many more than two classes.
Is there a workaround for this?
You most probably already have an included class that goes by that name, the likely problem is that you have a class by that name in another namespace (and said class has been included). You can use both classes (from different namespaces with the same name) by aliasing or using fully qualified names of the classes.
Read this properly.
I'm new to Laravel and using PHP namespaces in general. I didn't run into any problems until I decided to make a model named File. How would I go about namespacing correctly so I can use my File model class?
The files are app/controllers/FilesController.php and app/models/File.php. I am trying to make a new File in FilesController.php.
Namespacing is pretty easy once you get that hang of it.
Take the following example:
app/models/File.php
namespace App\Models;
class File {
public function someMethodThatGetsFiles()
{
}
}
app/controllers/FileController.php
namespace App\Controllers;
use App\Models\File;
class FileController {
public function someMethod()
{
$file = new File();
}
}
Declare the Namespace:
namespace App\Controllers;
Remember, once you've put a class in a Namespace to access any of PHP's built in classes you need to call them from the Root Namespace. e.g: $stdClass = new stdClass(); will become $stdClass = new \stdClass(); (see the \)
"Import" other Namespaces:
use App\Models\File;
This Allows you to then use the File class without the Namespace prefix.
Alternatively you can just call:
$file = new App\Models\File();
But it's best practice to put it at the top in a use statement as you can then see all the file's dependencies without having to scan the code.
Once that's done you need to them run composer dump-autoload to update Composer's autoload function to take into account your newly added Classes.
Remember, if you want to access the FileController via a URL then you'll need to define a route and specify the full namespace like so:
Route::get('file', 'App\\Controllers\\FileController#someMethod');
Which will direct all GET /file requests to the controller's someMethod()
Take a look at the PHP documentation on Namespaces and Nettut's is always a good resource with this article
first, load your class with:
$ composer dump-autoload
then
$file = new File;
// your stuff like:
$file->name = 'thename';
$file->active = true;
$file->save();
Section: Insert, Update, Delete on Laravel 4 Eloquent's doc
To namespace your model, at the top of your model class right after the opening
Then when you call from controllers you will call new Whatever\Model;
You probably have to do a dump-autoload with composer the first time around.
have a look to it.. hopefully will clear your query....
<?php
namespace app\controllers;
use yii\web\Controller;
use app\models\users;
class UserController extends Controller{
public function actionIndex()
{
echo "working on .....";
}
}
Namespaces are defined at the top of PHP classes right after the opening php script tag like this:
<?php
namespace MyNameSpace;
When you then want to use the namespaced class in some other class, you define it like this:
new MyNameSpace\PhpClass;
or import it at the top of the file (after namespaces if present) like this:
<?php
//namespace
use MyNameSpace\MyPHPClass;
//then later on the code you can instantiate the class normally
$myphpclass = new MyPHPClass();
In Laravel namespaces can be defined anywhere composer can autoload them, I'd recommend defining namespaces within the app directory. So you can define a namespace like Utils for holding Utility classes by creating a Utils directory in the app directory, creating our utility classes and defining the namespace as we did above.
Afterwards you have run the command to ask composer to autoload classes:
$ composer dump-autoload
When using the pseudo namespacing pattern of PEAR and Zend, it is common to come across class heirarchies that look like this:
Zend/
Db.php
Db/
Expr.php
Where DB.php contains a class named Zend_Db and Expr.php contains a class named Zend_Db_Expr. However, when you try to convert the old 5.2 psuedo namespacing into PHP 5.3 namespacing you are presented with a case where a namespace and a class share a name. Since the use operator can import either a namespace or a classname this leads to ambiguity.
Here's an example of an app I'm working on converting:
App/
Core.php
Core/
Autoloader.php
Here the base directory and namespace are App. In the top level of the name space is a Core class:
namespace App;
class Core { }
In the Core directory are various other core classes, some of which use the main Core. Under the pseudo namespacing pattern, this isn't a problem. But in the real namespacing pattern it creates this situation:
namespace App\Core;
use App\Core as Core; // What is this importing? Namespace or class?
class Autoloader {
public function __construct(Core $core) {}
}
Is this defined? What is actually imported here?
Simply both. It is not real import, just a hint for compiler, that every encounter of this alias in class related operations should be expanded to this declaration. In php namespace is just part of class so just think of it like this
$alias = 'Zend_Db';
$zendDB = new $alias;
$aliasExpr = $alias . '_Expr';
$expr = new $aliasExpr;