PHP Conference Nagoya 2025

    Namespaces und dynamische Sprachfeatures

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

    Die Implementierung von Namespaces in PHP ist stark von seinen Eigenschaften als Programmiersprache mit dynamischen Features beeinflusst. Man kann also den folgenden Code in Code mit Namespaces umformen:

    Beispiel #1 Dynamischer Zugriff auf Elemente

    example1.php:

    <?php
    class classname
    {
    function     __construct()
    {
    echo     __METHOD__,"\n";
    }
    }
    function     funcname()
    {
    echo     __FUNCTION__,"\n";
    }
    const     constname = "global";

    $a = 'classname';
    $obj = new $a; // gibt classname::__construct aus
    $b = 'funcname';
    $b(); // gibt funcname aus
    echo constant('constname'), "\n"; // gibt global aus
    ?>
    Um diesen Code umzuformen, muss man den vollständig qualifizierten Namen (Klassenname mit namespace-Präfix) verwenden. Beachten Sie, dass, weil es keinen Unterschied zwischen einem qualifizierten und vollständig qualifizierten Namen innerhalb eines dynamischen Klassen-, Funktions- oder Konstantennamen gibt, der führende Backslash nicht notwendigerweise angegeben werden muss.

    Beispiel #2 Dynamischer Zugriff auf Elemente mit Namespace

    <?php
    namespace namespacename;
    class     classname
    {
    function     __construct()
    {
    echo     __METHOD__,"\n";
    }
    }
    function     funcname()
    {
    echo     __FUNCTION__,"\n";
    }
    const     constname = "namespaced";

    /* Wenn man doppelte Anführungszeichen verwendet,
    muss "\\namespacename\\classname" verwendet werden */
    $a = '\namespacename\classname';
    $obj = new $a; // gibt namespacename\classname::__construct aus
    $a = 'namespacename\classname';
    $obj = new $a; // gibt ebenfalls namespacename\classname::__construct aus
    $b = 'namespacename\funcname';
    $b(); // gibt namespacename\funcname aus
    $b = '\namespacename\funcname';
    $b(); // gibt ebenfalls namespacename\funcname aus
    echo constant('\namespacename\constname'), "\n"; // gibt namespaced aus
    echo constant('namespacename\constname'), "\n"; // gibt ebenfalls namespaced aus
    ?>

    Bitte lesen Sie auch den Hinweis zum Escaping von Namespacenamen in Strings.

    Found A Problem?

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    User Contributed Notes 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
    3 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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