Namespaces e recursos de linguagem dinâmica

    (PHP 5 >= 5.3.0, PHP 7, PHP 8)

    A implementação de namespaces do PHP é influenciada por sua natureza dinâmica como linguagem de programação. Assim, para converter código como o exemplo a seguir em código com namespace:

    Exemplo #1 Acessando elementos dinamicamente

    exemplo1.php:

    <?php
    class nomedaclasse
    {
    function     __construct()
    {
    echo     __METHOD__,"\n";
    }
    }
    function     nomedafuncao()
    {
    echo     __FUNCTION__,"\n";
    }
    const     nomedaconstante = "global";

    $a = 'nomedaclasse';
    $obj = new $a; // imprime nomedaclasse::__construct
    $b = 'nomedafuncao';
    $b(); // imprime nomedafuncao
    echo constant('nomedaconstante'), "\n"; // imprime global
    ?>
    Deve-se usar o nome totalmente qualificado (nome da classe prefixado com o namespace). Observe que, como não há diferença entre um nome qualificado e um nome totalmente qualificado em um nome de classe, nome de função ou nome de constante dinâmico, a barra invertida inicial não é necessária.

    Exemplo #2 Acessando dinamicamente elementos com namespace

    <?php
    namespace nomedonamespace;
    class     nomedaclasse
    {
    function     __construct()
    {
    echo     __METHOD__,"\n";
    }
    }
    function     nomedafuncao()
    {
    echo     __FUNCTION__,"\n";
    }
    const     nomedaconstante = "comnamespace";

    /* note que, ao usar aspas duplas, "\\nomedonamespace\\nomedaclasse" deve ser usado */
    $a = '\nomedonamespace\nomedaclasse';
    $obj = new $a; // imprime nomedonamespace\nomedaclasse::__construct
    $a = 'nomedonamespace\nomedaclasse';
    $obj = new $a; // também imprime nomedonamespace\nomedaclasse::__construct
    $b = 'nomedonamespace\nomedafuncao';
    $b(); // imprime nomedonamespace\nomedafuncao
    $b = '\nomedonamespace\nomedafuncao';
    $b(); // também imprime nomedonamespace\nomedafuncao
    echo constant('\nomedonamespace\nomedaconstante'), "\n"; // imprime comnamespace
    echo constant('nomedonamespace\nomedaconstante'), "\n"; // também imprime comnamespace
    ?>

    Certifique-se de ler a nota sobre escapamento de namespaces em strings.

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    Notas Enviadas por Usuários (em inglês) 8 notes

    up
    75
    Alexander Kirk
    13 years ago
    When extending a class from another namespace that should instantiate a class from within the current namespace, you need to pass on the namespace.

    <?php // File1.php
    namespace foo;
    class     A {
    public function     factory() {
    return new     C;
    }
    }
    class     C {
    public function     tell() {
    echo     "foo";
    }
    }
    ?>

    <?php // File2.php
    namespace bar;
    class     B extends \foo\A {}
    class     C {
    public function     tell() {
    echo     "bar";
    }
    }
    ?>

    <?php
    include "File1.php";
    include     "File2.php";
    $b = new bar\B;
    $c = $b->factory();
    $c->tell(); // "foo" but you want "bar"
    ?>

    You need to do it like this:

    When extending a class from another namespace that should instantiate a class from within the current namespace, you need to pass on the namespace.

    <?php // File1.php
    namespace foo;
    class     A {
    protected     $namespace = __NAMESPACE__;
    public function     factory() {
    $c = $this->namespace . '\C';
    return new     $c;
    }
    }
    class     C {
    public function     tell() {
    echo     "foo";
    }
    }
    ?>

    <?php // File2.php
    namespace bar;
    class     B extends \foo\A {
    protected     $namespace = __NAMESPACE__;
    }
    class     C {
    public function     tell() {
    echo     "bar";
    }
    }
    ?>

    <?php
    include "File1.php";
    include     "File2.php";
    $b = new bar\B;
    $c = $b->factory();
    $c->tell(); // "bar"
    ?>

    (it seems that the namespace-backslashes are stripped from the source code in the preview, maybe it works in the main view. If not: fooA was written as \foo\A and barB as bar\B)
    up
    11
    Daan
    5 years ago
    Important to know is that you need to use the *fully qualified name* in a dynamic class name. Here is an example that emphasizes the difference between a dynamic class name and a normal class name.

    <?php
    namespace namespacename\foo;

    class     classname
    {
    function     __construct()
    {
    echo     'bar';
    }
    }

    $a = '\namespacename\foo\classname'; // Works, is fully qualified name
    $b = 'namespacename\foo\classname'; // Works, is treated as it was with a prefixed "\"
    $c = 'foo\classname'; // Will not work, it should be the fully qualified name

    // Use dynamic class name
    new $a; // bar
    new $b; // bar
    new $c; // [500]: / - Uncaught Error: Class 'foo\classname' not found in

    // Use normal class name
    new \namespacename\foo\classname; // bar
    new namespacename\foo\classname; // [500]: / - Uncaught Error: Class 'namespacename\foo\namespacename\foo\classname' not found
    new foo\classname; // [500]: / - Uncaught Error: Class 'namespacename\foo\foo\classname' not found
    up
    6
    museyib dot e at gmail dot com
    5 years ago
    Be careful when using dynamic accessing namespaced elements. If you use double-quote backslashes will be parsed as escape character.

    <?php
    $a    ="\namespacename\classname"; //Invalid use and Fatal error.
    $a="\\namespacename\\classname"; //Valid use.
    $a='\namespacename\classname'; //Valid use.
    ?>
    up
    15
    guilhermeblanco at php dot net
    15 years ago
    Please be aware of FQCN (Full Qualified Class Name) point.
    Many people will have troubles with this:

    <?php

    // File1.php
    namespace foo;

    class     Bar { ... }

    function     factory($class) {
    return new     $class;
    }

    // File2.php
    $bar = \foo\factory('Bar'); // Will try to instantiate \Bar, not \foo\Bar

    ?>

    To fix that, and also incorporate a 2 step namespace resolution, you can check for \ as first char of $class, and if not present, build manually the FQCN:

    <?php

    // File1.php
    namespace foo;

    function     factory($class) {
    if (    $class[0] != '\\') {
    echo     '->';
    $class = '\\' . __NAMESPACE__ . '\\' . $class;
    }

    return new     $class();
    }

    // File2.php
    $bar = \foo\factory('Bar'); // Will correctly instantiate \foo\Bar

    $bar2 = \foo\factory('\anotherfoo\Bar'); // Wil correctly instantiate \anotherfoo\Bar

    ?>
    up
    6
    akhoondi+php at gmail dot com
    11 years ago
    It might make it more clear if said this way:

    One must note that when using a dynamic class name, function name or constant name, the "current namespace", as in http://www.php.net/manual/en/language.namespaces.basics.php is global namespace.

    One situation that dynamic class names are used is in 'factory' pattern. Thus, add the desired namespace of your target class before the variable name.

    namespaced.php
    <?php
    // namespaced.php
    namespace Mypackage;
    class     Foo {
    public function     factory($name, $global = FALSE)
    {
    if (    $global)
    $class = $name;
    else
    $class = 'Mypackage\\' . $name;
    return new     $class;
    }
    }

    class     A {
    function     __construct()
    {
    echo     __METHOD__ . "<br />\n";
    }
    }
    class     B {
    function     __construct()
    {
    echo     __METHOD__ . "<br />\n";
    }
    }
    ?>

    global.php
    <?php
    // global.php
    class A {
    function     __construct()
    {
    echo     __METHOD__;
    }
    }
    ?>

    index.php
    <?php
    // index.php
    namespace Mypackage;
    include(    'namespaced.php');
    include(    'global.php');

    $foo = new Foo();

    $a = $foo->factory('A'); // Mypackage\A::__construct
    $b = $foo->factory('B'); // Mypackage\B::__construct

    $a2 = $foo->factory('A',TRUE); // A::__construct
    $b2 = $foo->factory('B',TRUE); // Will produce : Fatal error: Class 'B' not found in ...namespaced.php on line ...
    ?>
    up
    2
    m dot mannes at gmail dot com
    7 years ago
    Case you are trying call a static method that's the way to go:

    <?php
    class myClass
    {
    public static function     myMethod()
    {
    return     "You did it!\n";
    }
    }

    $foo = "myClass";
    $bar = "myMethod";

    echo     $foo::$bar(); // prints "You did it!";
    ?>
    up
    2
    anisgazig at gmail dot com
    2 years ago
    <?php

    //single or double quotes with single or double backslash in dynamic namespace class.

    namespace Country_Name{
    class     Mexico{
    function     __construct(){
    echo     __METHOD__,"<br>";
    }
    }

    $a = 'Country_Name\Mexico';//Country_Name\Mexico::__construct
    $a = "Country_Name\Mexico";
    //Country_Name\Mexico::__construct
    $a = '\Country_Name\Mexico';
    //Country_Name\Mexico::__construct
    $a = "\Country_Name\Mexico";
    //Country_Name\Mexico::__construct
    $a = "\\Country_Name\\Mexico";
    //Country_Name\Mexico::__construct
    $o = new $a;

    }

    /* if your namespace name or class name start with lowercase n then you should be alart about the use of single or double quotes with backslash */

    namespace name_of_country{
    class     Japan{
    function     __construct()
    {
    echo     __METHOD__,"<br>";
    }

    }

    $a = 'name_of_country\Japan';
    //name_of_country\Japan::__construct
    $a = "name_of_country\Japan";
    //name_of_country\Japan::__construct
    $a = '\name_of_country\Japan';
    //name_of_country\Japan::__construct
    //$a = "\name_of_country\Japan";
    //Fatal error: Uncaught Error: Class ' ame_of_country\Japan' not found
    //In this statement "\name_of_country\Japan" means -first letter n with "\ == new line("\n). for fix it we can use double back slash or single quotes with single backslash.
    $a = "\\name_of_country\\Japan";
    //name_of_country\Japan::__construct
    $o = new $a;
    }

    namespace     Country_Name{
    class     name{
    function     __construct(){
    echo     __METHOD__,"<br>";
    }
    }

    $a = 'Country_Name\name';
    //Country_Name\Norway::__construct
    $a = "Country_Name\name";
    //Country_Name\Norway::__construct
    $a = '\Country_Name\name';
    //Country_Name\Norway::__construct
    //$a = "\Country_Name\name";
    //Fatal error: Uncaught Error: Class '\Country_Name ame' not found

    //In this statement "\Country_Name\name" at class name's first letter n with "\ == new line("\n). for fix it we can use double back slash or single quotes with single backslash
    $a = "\\Country_Name\\name";
    //Country_Name\name::__construct
    $o = new $a;

    }

    //"\n == new line are case insensitive so "\N could not affected

    ?>
    up
    1
    scott at intothewild dot ca
    15 years ago
    as noted by guilhermeblanco at php dot net,

    <?php

    // fact.php

    namespace foo;

    class     fact {

    public function     create($class) {
    return new     $class();
    }
    }

    ?>

    <?php

    // bar.php

    namespace foo;

    class     bar {
    ...
    }

    ?>

    <?php

    // index.php

    namespace foo;

    include(    'fact.php');

    $foofact = new fact();
    $bar = $foofact->create('bar'); // attempts to create \bar
    // even though foofact and
    // bar reside in \foo

    ?>
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