I have a project that includes a PHP file that defines a class named Response and I'm auto-loading Guzzle via composer (which also defines it's own Response class)
The question is do I run the risk that PHP might get confused as to which Response class it should use? If I use Response by itself I don't run the risk that it will somehow fetch the Guzzle class unless I do a use statement right?
I think it's fine unless I do the use statement but I just wanna be sure.
You don't have any risk of conflicts, that's why PHP Namespaces exist!
Although the short name of both classes are the same, their FQCN (Fully Qualified Class Name) are different, so there's no chance of getting confused.
Guzzle's Response class is:
\Guzzle\Http\Message\Response
Your class would be something like:
\MyApp\Response
(or just \Response if you didn't set a namespace for it, which is a very bad practise!)
You can use either of them in your classes, or even both, as long as you set an alias for one of them:
use \MyApp\Response;
use \Guzzle\Http\Message\Response as GuzzleResponse;
// ...
$myResponse = new Response();
$guzzleResponse = new GuzzleResponse();
As long as the classes got individual namespaces.
Composer (and its autolaoder) will resolve the namespace with the classname.
So if you use your own class ("Class1") and the foreign class ("Class2") with the same classname ("Classname") with different namespaces e.g.
namespace \Me\MyPackage\Classname
namespace \Guzzle\Classname
you will be fine. If you wanna create an instance of one of the classes just type the full qualified namespace e.g.
$var1 = new \Me\MyPackage\Classname();
Related
I used use the keyword "use" generally above the class definition. Like this:
<?php
namespace suites\plugins\content\agpaypal;
use \Codeception\Util\Fixtures;
use \Codeception\Verify;
use \Codeception\Specify;
class agpaypalTest extends \Codeception\Test\Unit
{
protected $tester;
...
But now I realised, that I have to put the line for the trait Specify into the class definition. Like this:
<?php
namespace suites\plugins\content\agpaypal;
use \Codeception\Util\Fixtures;
use \Codeception\Verify;
class agpaypalTest extends \Codeception\Test\Unit
{
use \Codeception\Specify;
protected $tester;
...
I think it is because the package \Codeception\Specify; is a trait. But I do not understand why I couldn't reuse this trait when I set the line
use \Codeception\Specify;
before the class definition?
I would be happy if someone could point me to a hint or an explanaiton that explains to me where I should use the keyword "use" the best.
In PHP, the keyword use is used in 3 cases:
As class name alias - simply declares short name for a class (must be declared outside of the class definition)
(manual: Using namespaces: Aliasing/Importing )
To add a trait to a class (must be declared inside (at the top) of the class definition)
(manual: Traits)
In anonymous function definition to pass variables inside the function
(manual: Anonymous functions)
You can not import class with use keyword. You have to use include/require statement. Even if you use some php auto loader, still autoloader will have to use either include or require internally.
The Purpose of use keyword:
Consider a case where you have two classes with same name; you'll find it strange, but when you are working with big MVC structure, this happens. So if you have two classes with same name, put them in different name spaces. Now consider when your auto loader is loading both classes (does by require), and you are about to use object of class. In this case, the compiler will get confused which class object to load among two. To help the compiler make a decision, you can use the use statement so that it can make a decision which one is going to be used on.
Here refer this
How does the keyword 'use' work
use is basically including a class in the file to use it.
There are two ways to include a class file in another file.
The most general is require or include method. Another method is using composer. Assume this Directory Structure
Project
|
|--- Folder A
| |
| |---UserRegistration.php
|
|---Example
|
|--TestUserRegistration.php
In Folder A there is UserRegistartion.php and you want to use the code in TestUserRegistration.php In UserRegistration.php It can be class, trait or Interface
Method 1.
In TestUserRegisteration.php you can include or require file UserRegistartion.php
and use it
Method 2
Using Composer. In UserRegistration.php you define namespace FolderA; as the first line of code. Then write your code as you do. So Now you want to use this file in TestUserRegistration.php you do
include vendor/autoload.php;
use FolderA\UserRegistration;
Which one is better and why?
Method 2 using composer is the best method. In method 1 wherever you want to include UserRegistration you have to find the relative path to UserRegistration file. So lets assume some day you need to change the directory structure your application will break as the relative path you had provided now it does'nt exist.
But in Method 2 you always use the namespace you provided \ The filename instead of where you want to use. So even you change the directory structure you don't have to got all codes and modify the path. It will work as it was.
To know more study about how to use namespace and composer.
I want to write a small add on for an existing CMS. To do so, I need to extend a class from that CMS' code.
My code will be written inside its own namespace, while the CMS' code does not use namespacing, which basically means it exists inside the global namespace.
Inside my code, I create a new stdClass:
$var = new stdClass();
With that code is place, it always produces a fatal error:
Fatal error: Class 'MyNamespace\something\StdClass' not found in /some/rather/long/path/to/class.php on line 123
Creating the stdClass like this solves that problem:
$var = new \stdClass();
Since I am still pretty new to namespaces, I am not exactly sure what the problem here is?
My guess is that in the first example, the stdClass would be created in the namespace of my class. This actually means the constructor of a class called stdClass existing in my namespace would be called, but since that class does not exist, an error is thrown.
In the second example, I signalize that I want to instantiate the class called stdClass from the global namespace, which somehow suddenly makes sense.
If anyone could elaborate what is happening here I would be very happy.
You appear to understand the concept behind namespaces, and you are headed in the right direction on your analysis of what is happening.
When you are working inside of a namespace you are able to refer to names as unqualified, qualified, and fully qualified.
When you make a namespace you are telling PHP to organize (and resolve) the names of your classes, function, methods, etc. away from the same scope where the built-in PHP code lives along with any other code behind its own namespace. It is away to organize your code and avoid naming collisions among libraries and built-in PHP functions.
Here is a brief on how names get resolved:
If you are trying to resolve a name within the same namespace you can use the unqualified name. So for class \Foo\Bar\Baz you can use new Baz(); as long as you are in namespace \Foo\Bar.
If you are trying to resolve a name that is lower in the same parent namespace you can use the qualified name. So for class \Foo\Bar\Baz you would need to use new Bar\Baz(); if you were in namespace \Foo.
If you are trying to resolve a name that is not in your namespace or is in the global namespace (built-in PHP stuff) then you must use the fully qualified name. If you are in namespace \Foo\Bar and you want to make use of something like the mysqli class you would need to call it by its fully qualified name. e.g. new \mysqli() Your question above is a perfect example that illustrates this. Likewise, if you need to access a class in a totally different namespace you would also need the fully qualified name: new \Third\Party\AppClass();
So to summarize, you are right, the built-in stdClass does not exist in your namespace, therefore you need to access it by the fully qualified name. The reason you must do thing this way had to do with conforming to the rules PHP uses when resolving names.
If you ever need to find out what namespace you are in it will be in the __NAMESPACE__ constant.
In case you haven't already read it, here is the documentation on name resolution in PHP: http://php.net/manual/en/language.namespaces.rules.php
The code is evaluated in your namespace and stdClass doesn't exist there. You are effectively answering your own question with your guess.
Your guess is correct. Look at the comments in "class references" section of Example #1 on this page.
http://php.net/manual/en/language.namespaces.rules.php
I have a laravel 4.1 application, and I've created a folder in my app folder to store most of the logic.
/app/Acme/Models/
/app/Acme/Repositories/
these are the two main folders.
In my composer.json I have this in the auto load, and done a dump run.
"psr-4" : {
"Acme\\" : "app/Acme"
}
However I am getting, what I think are silly issues. For example my Acme/Models/Task.php has the following
<?php
namespace Acme\Models;
class Task extends \Eloquent {
public function job()
{
return $this->belongsTo('Job');
}
}
however when I run this, I get an error
Fatal error: Class 'Task' not found in vendor/laravel/framework/src/Illuminate/Database/Eloquent/Model.php on line 780
In my Job.php I have the same namespace at the top of the file....
Must I manually import/use object which are in the same name space?
use Acme/Models/Job as Job; ? it seems like such a duplicate...
And in my Repositories folder when I set a namespace of namespace Acme/Repositories;, must I use items like
use Acme\Models\Job as Job;
I is, a bit lost!
Namespaces are relative so you do not need to add use to directly reference classes within the same namespace.
The error you are getting is because you need to fully qualify relationships to namespaced models so eloquent knows where to find them eg
$this->belongsTo('\Acme\Models\Jobs');
In the case of your repository namespace, you will need to add a use statement in your file as you suggested, or reference the fully qualified namespace eg new \Acme\Models\Job();
On a side note, I know PHPStorm (and I'm sure other IDEs) will inject the namespaces for you which is super useful and saves you having to write use or the full namespace out every time you reference a class - worth checking out.
Edit: Sorry, I didn't read the question properly first time around - updated my answer.
I'm trying to rewrite an OO PHP site (that loosely follows an MVC structure) so it uses namespaces - and want to follow PSR-0.
In the current site I have a class (called APP) which is full of static methods that I call all over the place to handle things such as getting config data eg; APP::get_config('key').
Obviously with namespacing, I would need to call \TheNameSpace\App::get_config('key'). I use this class frequently, so want to avoid having to prefix the namespace every time I use it. I do call methods in it from within other classes, which would usually be under a sub-namespace - so changing the namespace at the top of the file won't really work.
So, I guess my question is, what is the easiest way to have a 'global' class with methods that I can call anywhere without having to prefix with the namespace each time?
namespace Foo;
use Bar;
Then you do not have to do \Bar\fn
So in your case:
namspace Foo;
use TheNameSpace\App;
App::get_config('blah')
Read the section in the php manual on using/aliasing namespaces.
http://www.php.net/manual/en/language.namespaces.importing.php
You can exclude the namespace by using "use". You can name it whatever you want.
use TheNamespace\App as App //You can name it anything here
App:config('key');
At top of your scripts add
use TheNameSpace\App as MyApp
for example. You can then use it like
app = new MyApp();
in your scripts. Of course you needn't to use an alias here. Just
use TheNameSpace\App
app = new App();
will work, too.
A global class that's implementing this one is bad style and you shouldn't do it like this:
class MyApp extends TheNameSpace\App { }
....
myApp = new MyApp();
I have a file with a class Resp. The path is:
C:\xampp\htdocs\One\Classes\Resp.php
And I have an index.php file in this directory:
C:\xampp\htdocs\Two\Http\index.php
In this index.php file I want to instantiate a class Resp.
$a = new Resp();
I know I can use require or include keywords to include the file with a class:
require("One\Classes\Resp.php"); // I've set the include_path correctly already ";C:\xampp\htdocs". It works.
$a = new Resp();
But I want to import classes without using require or include. I'm trying to understand how use keyword works. I tried theses steps but nothing works:
use One\Classes\Resp;
use xampp\htdocs\One\Classes\Resp;
use htdocs\One\Classes\Resp;
use One\Classes;
use htdocs\One\Classes; /* nothing works */
$a = new Resp();
It says:
Fatal error: Class 'One\Classes\Resp' not found in C:\xampp\htdocs\Two\Http\index.php
How does the keyword use work? Can I use it to import classes?
No, you can not import a class with the use keyword. You have to use include/require statement. Even if you use a PHP auto loader, still autoloader will have to use either include or require internally.
The Purpose of use keyword:
Consider a case where you have two classes with the same name; you'll find it strange, but when you are working with a big MVC structure, it happens. So if you have two classes with the same name, put them in different namespaces. Now consider when your auto loader is loading both classes (does by require), and you are about to use object of class. In this case, the compiler will get confused which class object to load among two. To help the compiler make a decision, you can use the use statement so that it can make a decision which one is going to be used on.
Nowadays major frameworks do use include or require via composer and psr
1) composer
2) PSR-4 autoloader
Going through them may help you further.
You can also use an alias to address an exact class. Suppose you've got two classes with the same name, say Mailer with two different namespaces:
namespace SMTP;
class Mailer{}
and
namespace Mailgun;
class Mailer{}
And if you want to use both Mailer classes at the same time then you can use an alias.
use SMTP\Mailer as SMTPMailer;
use Mailgun\Mailer as MailgunMailer;
Later in your code if you want to access those class objects then you can do the following:
$smtp_mailer = new SMTPMailer;
$mailgun_mailer = new MailgunMailer;
It will reference the original class.
Some may get confused that then of there are not Similar class names then there is no use of use keyword. Well, you can use __autoload($class) function which will be called automatically when use statement gets executed with the class to be used as an argument and this can help you to load the class at run-time on the fly as and when needed.
Refer this answer to know more about class autoloader.
use doesn't include anything. It just imports the specified namespace (or class) to the current scope
If you want the classes to be autoloaded - read about autoloading
Don’t overthink what a Namespace is.
Namespace is basically just a Class prefix (like directory in Operating System) to ensure the Class path uniqueness.
Also just to make things clear, the use statement is not doing anything only aliasing your Namespaces so you can use shortcuts or include Classes with the same name but different Namespace in the same file.
E.g:
// You can do this at the top of your Class
use Symfony\Component\Debug\Debug;
if ($_SERVER['APP_DEBUG']) {
// So you can utilize the Debug class it in an elegant way
Debug::enable();
// Instead of this ugly one
// \Symfony\Component\Debug\Debug::enable();
}
If you want to know how PHP Namespaces and autoloading (the old way as well as the new way with Composer) works, you can read the blog post I just wrote on this topic: https://enterprise-level-php.com/2017/12/25/the-magic-behind-autoloading-php-files-using-composer.html
You'll have to include/require the class anyway, otherwise PHP won't know about the namespace.
You don't necessary have to do it in the same file though. You can do it in a bootstrap file for example. (or use an autoloader, but that's not the topic actually)
The issue is most likely you will need to use an auto loader that will take the name of the class (break by '\' in this case) and map it to a directory structure.
You can check out this article on the autoloading functionality of PHP. There are many implementations of this type of functionality in frameworks already.
I've actually implemented one before. Here's a link.
I agree with Green, Symfony needs namespace, so why not use them ?
This is how an example controller class starts:
namespace Acme\DemoBundle\Controller;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
class WelcomeController extends Controller { ... }
Can I use it to import classes?
You can't do it like that besides the examples above. You can also use the keyword use inside classes to import traits, like this:
trait Stuff {
private $baz = 'baz';
public function bar() {
return $this->baz;
}
}
class Cls {
use Stuff; // import traits like this
}
$foo = new Cls;
echo $foo->bar(); // spits out 'baz'
The use keyword is for aliasing in PHP and it does not import the classes. This really helps
1) When you have classes with same name in different namespaces
2) Avoid using really long class name over and over again.
Using the keyword "use" is for shortening namespace literals. You can use both with aliasing and without it. Without aliasing you must use last part of full namespace.
<?php
use foo\bar\lastPart;
$obj=new lastPart\AnyClass(); //If there's not the line above, a fatal error will be encountered.
?>
Namespace is use to define the path to a specific file containing a class e.g.
namespace album/className;
class className{
//enter class properties and methods here
}
You can then include this specific class into another php file by using the keyword "use" like this:
use album/className;
class album extends classname {
//enter class properties and methods
}
NOTE: Do not use the path to the file containing the class to be implements, extends of use to instantiate an object but only use the namespace.