PHP: Namespaces in a single file with a global namespace - php

I have a file that require()'s a namespace, as such:
<?php
require_once('Beer.php'); // This file contains the Beer namespace
$foo = new Beer\Carlsburg();
?>
I would like to put the Beer namespace directly in the same file, like this (unworking) example:
<?php
namespace Beer {
class Carlsburg {}
}
$foo = new Beer\Carlsburg();
?>
However, the PHP interpreter complains that No code may exist outside of namespace. I can therefore wrap $foo declaration in a namespace, but then I must also wrap Beer in that namespace to access it! Here is a working example of what I am trying to avoid:
<?php
namespace Main\Beer {
class Carlsburg {}
}
namespace Main {
$foo = new Beer\Carlsburg();
}
?>
Is there any way to include the code for the Beer namespace in the file, yet not wrap the $foo declaration in its own namespace (leave it in the global namespace)?
Thanks.


You should use the global namespace :
<?php
namespace Beer {
class Carlsburg {}
}
namespace { // global code
$foo = new Beer\Carlsburg();
}
?>
See here -> http://php.net/manual/en/language.namespaces.definitionmultiple.php


Try this
namespace Beer {
class Carlsburg {}
}
//global scope
namespace {
$foo = new Beer\Carlsburg();
}
As per example #3 in Defining multiple namespaces in the same file


Try placing a backslash before the namespace name:
$beer = new \Beer\Carlsberg();
The initial backslash is translated to "global namespace". If you do not put the leading backslash, the class name is translated to the current namespace.


Just write it, it has no "name":
<?php
namespace Main\Beer {
class Carlsburg {}
}
namespace {
use Main\Beer;
$foo = new Beer\Carlsburg();
}
?>
Demo and see Defining multiple namespaces in the same fileDocs.

Related

Overriding PHP namespace for a specific class

If I have a PHP namespace as follows:
<?php
namespace A\B;
$test = new MyClass();
it seems every time I create a new instance of class, the name is prefixed with the namespace, e.g.
$test = new A\B\MyClass();
What happens if I don't want to use a namespace for another class, e.g. I want to call
$test = C\D\AnotherClass();
Currently this becomes:
$test = A\B\C\D\AnotherClass();
which results in an error.

For php7 you can wrap code in a namespace then call it with use in the global namespace declared using namespace with no name per the example below.
namespace FRED\WILMA\BAMBAM {
const FW = 2;
function fw() {
echo "<br>fw<br>";
}
}
namespace BARNEY\BETTY { // Create Namespace and add function
function bb() {
echo "<br>bb<br>";
}
}
namespace { // use global namespace
use FRED\WILMA\BAMBAM;
use BARNEY\BETTY;
BETTY\bb();
BAMBAM\fw();
}

Can the namespace statement be "included" in PHP?

I have a file which looks like this:
<?php
namespace n;
f(); // this calls n\f()
PHP statements can be included by using the include function, however, this doesn't work for the namespace statement:
<?php
include('example_include_namespace_statement.php')
f(); // this doesn't call n\f()
example_include_namespace_statement.php:
<?php
namespace n;
Is there any way to "include" the namespace statement in PHP?
Or is it simply not possible?

You have to specify the namespace when calling a function that is in a namespace, like so:
// test.php
namespace test;
function doesItWork() {
return 'It Works!';
}
// index.php
include 'test.php';
echo test\doesItWork();
If you want to use a function that is defined in a namespace that is outside the current namespace then you must prefix it with the namespace character, like so:
echo \test\doesItWork();
You should read the documentation on namespaces.

PHP namespace and global autoload single file

This is an experiment with PHP namespaces / autoload in a single file.
namespace trust;
class trust_network{
public function __construct(){
print "SUP";
}
}
namespace trust2;
$trust = new \trust\trust_network(); $do = new \test();
function __autoload($class){
require($class.".php");
print $class;
}
So under namespace trust2, I'm calling "\test" - aka I'd like to autoload that class from an external file on a global base. What I wrote does not work. I know that I've got __autoload under a namespace, but how do I declare that on a global basis? Can't include before namespace declaration.

For multiple namespaces in one file you should use the curly bracket syntax:
namespace n1 {
...
}
namespace n2 {
...
}
namespace {
...
}
In the last block you can declare functions in the global namespace. Reference: http://www.php.net/manual/en/language.namespaces.definitionmultiple.php

Autoload is usually so that you can put one class per file. Therefore, you should have the following layout
/index.php
function __autoload($class){
// You may need to convert backslashes in $class to forward slashes
// and strip the first slash, we'll leave the
require($class.".php");
// debug-only: print $class;
}
// Calling new here triggers __autoload to be called
$trust = new \trust\trust_network();
$do = new \test();
/trust/trust_network.php
namespace trust;
class trust_network{
public function __construct(){
print "TRUST_NETWORK";
}
}
/test.php
class test() {
public function __construct(){
print "TEST";
}
}
Note that you should use spl_autoload_register instead since it allows multiple systems to hook in their own autoload behavior. As of PHP 5.3, you can do the following
spl_autoload_register(function ($class) {
require($class.".php");
});

Cannot find Class with PHP Namespace

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

Can I rename a namespace in PHP?

In C++ one can do this:
namespace qux = std::foo::bar::baz;
qux::CFoo BAR;
Can one do such a thing in PHP?

You can do this :
namespace foo\bar\baz;
use foo\bar\baz as renamed;
new renamed\cFoo(); // Points to foo\bar\baz\cFoo()
See the documentation for further details.

Namespaces may be aliased (docs).
The general idea is use … as …; as shown below.
use std\foo\bar\baz as qux;
qux\CFoo();
And here's a try-this-at-home example:
<?php
namespace std\foo\bar\baz {
function CFoo() {
echo 'hello, world';
}
}
namespace {
use std\foo\bar\baz as qux;
qux\CFoo();
}
?>

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