I'm trying to run this code on the same file:
namespace Foo1\Bar\SubBar;
class SubBarClass {
public function __construct() {
echo 'From Foo1';
}
}
namespace Foo2\Bar\SubBar;
class SubBarClass {
public function __construct() {
echo 'From Foo2';
}
}
use Foo1\Bar\SubBar;
$foo1 = new SubBarClass;
use Foo2\Bar\SubBar;
$foo2 = new SubBarClass;
The ideia is to change namespaces and echo the related value.
But it's returning the following error:
( ! ) Fatal error: Cannot use Foo2\Bar\SubBar as SubBar because the name is already in use in C:\wamp\www\xxx\namespaces.php on line 30
Line 30: use Foo2\Bar\SubBar;
How can I interchange namespaces on the same file?
Thks!
use keyword is used to import that namespace to be accessed in your current file scope. It does not act as a namespace "instance constructor".
You're current under Foo2\Bar\SubBar namespace. Like a directory of classes, while you're here, you should access other namespaces from the root (\):
$foo2 = new SubBarClass;
$foo1 = new \Foo1\Bar\SubBar\SubBarClass;
There is no need to use use for those namespaces (although you can, specially when they share parent namespaces), they are already declared in the same file you're using them.
For more information about this, consider reading the manual, where it describes using multiple namespaces in the same file.
This happens because the last defined namespace is the one currently active.
So, when I type:
use Foo1\Bar\SubBar;
I'm still on the last defined namespace: Foo2\Bar\SubBar.
Hence, when I type:
use Foo2\Bar\SubBar;
I'm trying to use the currently active namespace. That's why the Fatal error is returned.
On possible solution is:
namespace Foo1\Bar\SubBar;
class SubBarClass {
public function __construct() {
echo 'From Foo1';
}
}
namespace Foo2\Bar\SubBar;
class SubBarClass {
public function __construct() {
echo 'From Foo2';
}
}
use Foo1\Bar\SubBar;
$foo1 = new SubBar\SubBarClass;
echo '<br>';
$foo2 = new SubBarClass;
Cheers!
Related
I can not understand why php gives me an error
"Fatal error: Cannot declare class rex\builder\RexBuilder, because the
name is already in use in /var/www/site2.dev/App/rex/RexBuilder.php on
line 12"
RexBuilder static class, and it is called only 1 time.
I did a search on the project, no longer classes with the same name.
<?php
namespace rex\builder;
require_once 'Router.php';
use rex\router\Router;
error_reporting(E_ALL);
ini_set('display_errors', 1);
class RexBuilder {
public static function collector($array) {
$router = new Router();
foreach ($array as $key => $val) {
$router->get($val->getMethod(), $val->getInterfaces(), $val->getHandler());
}
$router->init();
}
}
?>
Call the class in index.php
RexBuilder::collector(array(
new BuildModel('POST', '/api/v1/user/register', new \api\register\Registration()),
new BuildModel('POST', '/api/v1/user/login', new \api\login\Login())));
More This class is not used
The error is thrown because of the use rex\router\Router; duplicate classes.
When you are writing use namespace.. it means you can go directly to that namespace like it is your current namespace
Lets take a look at the next code:
We'll create a file and declare that it belongs to namespace classes\a
//file: a.php
<?php
namespace classes\a;
class A{
}
now lets create another file b.php (and declare it belongs to namespace classes\b but it means nothing for the example)
namespace classes\b;
require_once "a.php";
use classes\a; //Notice that I'm using this namespace, it means I can use it directly
class A{
}
Generates the error
Fatal error: Cannot declare class classes\b\A because the name is already in use in
We have to solutions possible:
First: remove the use tag and write the namespace directly
class A{
function __constructor(){
$instance = new classes\a\A();
}
}
Second, give it alias
use classes\a as out_a;
class A{
function __constructor(){
$instance = new out_a\A();
}
}
For your code, just remove the use or give it an alias.
The problem is certainly because you include the RexBuilder.php file two times instead of one.
If you call the file by this way : include('RexBuilder.php'); or this way require('RexBuilder.php'); please change it by include_once('RexBuilder.php'); or require_once('RexBuilder.php'); which only allows ONE call of the file.
I have code - working correct (I do not have to inculde class ReflectionClass):
class Test
{
const TYPE_ONE = "Number one";
const TYPE_TWO = "Number two";
static function getConstants() {
$oClass = new ReflectionClass(__CLASS__);
return $oClass->getConstants();
}
}
foreach (Test::getConstants() as $kay => $val):
echo "$kay -- $val <br/>";
endforeach;
But, when i try use ReflectionClass in code Yii2 i gets the message
PHP Fatal Error – yii\base\ErrorException
Class 'common\models\ReflectionClass' not found
If there are any Reflection classes in the framework or a way to declare ReflectionClass in Yii2
Because yii2 use namespaces, when you call new ReflectionClass() php looking for this class in the namespace that you declare at the beginnig of a file, in your case its namespace common\models; To load php's classes you need to prepend their names with \. So to instantiate ReflectionClass you need to write new \ReflectionClass(__CLASS__). More in documentation
In my php file i am doing it this way
pagecontroller.php
include_once(RUDRA."/controller/AbstractTemplateController.php");
if (file_exists(get_include_path() . CONTROLLER_PATH . "/TemplateController.php" )) {
include_once (CONTROLLER_PATH . "/TemplateController.php");
} else {
include_once (RUDRA . "/controller/TemplateController.php");
}
in TemplateController.php a class named 'TemplateController extends AbstractTemplateController' is defined, if a developer has already defined a class TemplateController which also extends AbstractTemplateController then it will use that otherwise it will fallback to default definition.
then in other files i will simply use something like this
include_once("pagecontroller.php")
$c = new TemplateController();
is there any better way to do this?
since I am including two files AbstractTemplateController.php & TemplateController.php in both cases, I cpuld have written both class definitions in same file which would have saved one include(if there is no custom TemplateController.php)?
I tried writing AbstractTemplateController & TemplateController in one single file but if then developer has defined his own TemplateController it creates two classes with same name situation.
pupose is to have atleast one definition to be there, if customDefinition does not exists then only use default one. and this code is to be abstract.
in the beginning if CustomClass exists (in a specific folder) then that the definition to be used, else use default one (which is nothing but simply extends AbstractOne)
CONTROLLER_PATH . "/TemplateController.php"
class TemplateController extends AbstractTemplateController {
/* over-ridden method of AbstractTemplateController
*/
public function invoke($abc,$def){
echo $abc . " " .$def;
}
}
RUDRA . "/controller/TemplateController.php"
class TemplateController extends AbstractTemplateController {
// nothing at all this is simply to make sure TemplateController class is available
// for others to use.
}
Use namespaces and convention.
E.g. you could check if there's a TemplateController-class present that extends the AbstractTemplateController that's different from your namespace (As your implementation will be specific for your namespace), if there isn't ; fall back to your implementation of the TemplateController.
http://php.net/manual/en/language.namespaces.php
php provides a function for not letting you load/write a class more then once.
bool class_exists ( string $class_name );
example is :
<?php
function __autoload($class)
{
include($crigger_error("Unable to load class: $class", E_USER_WARNING);
}
}
if (class_exists('MyClass')) {
$myclass = new MyClass();
}lass . '.php');
// Check to see whether the include declared the class
if (!class_exists($class, false)) {
trigger_error("Unable to load class: $class", E_USER_WARNING);
}
}
if (class_exists('MyClass')) {
$myclass = new MyClass();
}
?>
in above example autoload is used, you could do it without autoload this way :
<?php
// Check that the class exists before trying to use it
if (class_exists('MyClass')) {
$myclass = new MyClass();
}
?>
still i am saying you better get habit of using namespaces. they are awesome and work every where.
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 $this->registry;
echo '<br>Registry was ran inside testcontroller.php<br>';
}
}
?>
and
<?php
namespace Library\Registry;
class Registry
{
function __construct()
{
return 'Registry.class.php Constructor was ran';
}
}
?>
First off, I believe you are using composer or composer is initialised in your project. If so, check composer.json file for your autoload, psr-4 definition. For example, if the root of your application is "App", then in your psr-4, you should be doing "autoload": { "psr-4": { "App\\": "./" } },
Furthermore, remember to clear composer cache and dump-autoload from the terminal as follows:
composer clear-cache
composer dump-autoload
I work a lot in PHP but I never really understand the namespace method in PHP. Can somebody help me here? I have read on php.net's website its not explained good enough, and I can't find examples on it.
I need to know how I can make code in sample version.
namespace: sample
class: sample_class_1
function: test_func_1
class: sample_class_2
function: test_func_2
function: test_func_3
Like this?
<?php
namespace sample
{
class Sample_class_1
{
public function test_func_1($text)
{
echo $text;
}
}
class Sample_class_2
{
public static function test_func_2()
{
$c = new Sample_class_1();
$c->test_func_1("func 2<br />");
}
public static function test_func_3()
{
$c = new Sample_class_1();
$c->test_func_1("func 3<br />");
}
}
}
// Now entering the root namespace...
// (You only need to do this if you've already used a different
// namespace in the same file)
namespace
{
// Directly addressing a class
$c = new sample\Sample_class_1();
$c->test_func_1("Hello world<br />");
// Directly addressing a class's static methods
sample\Sample_class_2::test_func_2();
// Importing a class into the current namespace
use sample\Sample_class_2;
sample\Sample_class_2::test_func_3();
}
// Now entering yet another namespace
namespace sample2
{
// Directly addressing a class
$c = new sample\Sample_class_1();
$c->test_func_1("Hello world<br />");
// Directly addressing a class's static methods
sample\Sample_class_2::test_func_2();
// Importing a class into the current namespace
use sample\Sample_class_2;
sample\Sample_class_2::test_func_3();
}
If you're in another file you don't need to call namespace { to enter the root namespace. So imagine the code below is another file "ns2.php" while the original code was in "ns1.php":
// Include the other file
include("ns1.php");
// No "namespace" directive was used, so we're in the root namespace.
// Directly addressing a class
$c = new sample\Sample_class_1();
$c->test_func_1("Hello world<br />");
// Directly addressing a class's static methods
sample\Sample_class_2::test_func_2();
// Importing a class into the current namespace
use sample\Sample_class_2;
sample\Sample_class_2::test_func_3();