As the title says, when I instantiate a class I get this message :
Fatal error: Class 'Envato\RecursiveIteratorIterator' not found in C:\Users\rgr\Apache\htdocs\Roland Groza [ 3.0 ]\class\envato\envato.php on line 359
You can view the class here : Class ;
I'm instantiating from another file :
require("envato.php");
$test = new Envato\EnvatoAPIWrapper();
echo "User Vitals : ".$test->get_user_vitals("chaoscod3r")."<br>";
The class is wrapped with a namespace, so that might have something to do with it, but I wasn't sure since it's been a few years since I haven't coded PHP. Hopefully someone has an idea what is it that I'm doing wrong :)
To access non-namespaced classes like the internal classes of PHP and SPL inside of a namespace you have to use the fully qualified class name like this:
new \RecursiveIteratorIterator();
or import it explicitly at the beginning:
use \RecursiveIteratorIterator;
and then use it normally like you do.
Add a use statement at the top of your namespace...
use \RecursiveIteratorIterator;
If you don't then PHP expects RecursiveIteratorIterator to exist within your current namespace, rather than in the global namespace (indicated by the leading \)
Related
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 just wanna use a namespace in another file, to use the class in it, but im too retarded to do this...
first file:
namespace fun;
use fun\kjr\trouble;
$trouble = new trouble('http://someURL');
second file:
namespace fun\kjr;
class trouble { ... }
This is the error i get:
Error: Fatal error: Class 'fun\kjr\trouble' not found in D:\wamp\www\fun\index.php on line 8
Where did I declare a wrong namespace? Oo
Greetings
I had to include my class aswell.
use namespace
only makes the class available in the actual context.
It does NOT include it, so you can cause it.
use
include_once('your.file')
to do that!
Thanks #Michael Berkowski for helping.
The problem (from what I can see) is you haven't specified an autoloader (e.g. with spl_autoload_register()) or haven't included the file in your code. The use statement will only find a class for you if you have an autoloader in place, otherwise you must include your code ahead of time with include/include_once or require/require once.
I would start with making an autoloader and registering it and then it may be wise to map your namespaces to directory structure as specified by PSR-0.
I have a lot of classes in multiples subfolders that I load using this autoloader :
spl_autoload_register(function ($class) {
$class = str_replace('\\', DIRECTORY_SEPARATOR, strtolower($class));
if(file_exists(FILES_PATH.'classes/'.$class.'.class.php')){
require_once(FILES_PATH.'classes/'.$class.'.class.php');
}
});
So if i do new Folder\subFolder\Myclass, it works.
The classes in folders are all in a namespace.
All these classes must use the database class, and the problem is here :
When the class is in a namespace and search the database class, it can't find it.
(DB class is in global namespace)
So I try to put "use BDD" (Bdd is the db class) and it still doesn't work, because Bdd is using PDO and so i must do "use bdd, pdo;" in EVERY classes of the project...
I find this stupid. Is this normal ?
Is there a better way to autoload, without using namespaces ?
It's pretty darn simple:
If you're in a namespace like so:
namespace Foo;
all names of all classes are resolved relative to that namespace. Any Bar will mean the class Foo\Bar, not the "global" Bar. If you want to refer in any way, shape or form to a class which is not in the same namespace, say Bar\Baz (class Baz from the namespace Bar), you have two choices:
use a fully qualified name for the class, e.g.:
\Bar\Baz
where the leading \ means the class name shall be resolved from the top namespace, not the current local one, or
if this is getting annoying to do every time, alias it using:
use Bar\Baz;
which is shorthand for
use Bar\Baz as Baz;
which means any time you use "Baz" in this namespace you mean the class Bar\Baz, not Foo\Bar\Baz.
Yes, this applies to each file individually. If you want to refer to PDO in some namespace in some file, you either have to write \PDO to make it resolve to the "global" PDO class or you write use PDO at the top of the file to make a convenience alias. That's how namespaces work.
This applies to all use cases of any class name:
new \PDO
\PDO::staticMethod()
\PDO::CONSTANT
You can explicitly say that BDD is in the global namespace by doing this in your code:
$foo = new \BDD();
Then you should not need to use it.
Moving answer from one of my comments ;)
Using use \Pdo; in BDD class fixed the world :)
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.
How can I "reset" a namespace to the global one? Given the following code:
namespace foo;
include 'myfile.php';
myfile.php will now try to load all of its classes in the foo namespace, even though its classes are in the global namespace. Now it wouldn't be a big deal to swap the order of those lines, but how would I deal with myfile.php having an autoloader? It will try to load the classes in namespace foo.
Namespaces work on a per-file basis. If myfile.php doesn't declare any namespace then everything in it will belong to the global namespace, regardless of the namespace in which include was used.
Long story short, namespace declarations only apply to their own file, not includes.
myfile.php should declare the namespace that it wants to be in. If you want to ensure that myfile.php loads in the global namespace, I believe you can put
namespace // empty namespace means global
{
}
around the body of myfile.php. Make sure that you use the braces, or else you'll redefine the rest of the file in which myfile.php is included to also be in the global namespace.
Maybe I didn't understand your question well ; in that case, could you provide an example of what you get, and what you'd expect ?
Are you sure it will try to load the functions from myfile.php in the foo namespace ?
Considering I have one file (temp-2.php) that looks like this :
<?php
namespace foo;
function my_function_this_file() {
var_dump(__FUNCTION__);
}
my_function_this_file();
include 'my_other_file.php';
my_function_the_other_file();
And another one (my_other_file.php) that looks like this :
<?php
function my_function_the_other_file() {
var_dump(__FUNCTION__);
}
When I call the first one from my browser, I get this output :
string 'foo\my_function_this_file' (length=25)
string 'my_function_the_other_file' (length=26)
This seems to indicate the second function is not inside any namespace, except the global one -- which corresponds to the fact that it's not declared in any namespace.
If I remember correctly, the "namespace" instruction is only valid for the file it is used in, and not for included files.
The Import names cannot conflict with classes defined in the same file page from the namespaces FAQ seems to indicate that too.
Hope this helps, and I understood the question correctly...
EDIT : btw, swapping the order of the to lines, like this :
<?php
include 'my_other_file.php';
namespace foo;
Wouldn't work : the "namespace" instruction must be the first one of the file : you'll get a Fatal Error, if you do that :
Fatal error: Namespace declaration statement has to be the very first statement in the script
I'd just like to chime in on this with my findings. Since PHP cannot handle exceptions in __toString methods, the solution is to do it yourself: something along the lines of
public function __toString()
{
try {
return $this->draw();
} catch ( \Exception $e ) {
die( \error::_getExceptionPage( $e ) );
}
}
However what becomes really weird, the namespace of the call to error::_getExceptionPage($e) becomes the same as the class in which __toString was defined - and I wasn't able to find a way around that except prefixing everything in my exception handling code with \, i.e. every call to any class in getExceptionPage must become \someClass::staticMethod().
Same goes for included files, everything - all code since the thrown exception becomes embedded in the namespace of the faulty class. Tested with PHP 5.3.17