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[edit] Last updated: Fri, 23 Mar 2012

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interface_exists

(PHP 5 >= 5.0.2)

interface_existsArayüz tanımlı mı diye bakar

Açıklama

bool interface_exists ( string $arayüz [, bool $özdevinimli_yükle = true ] )

Belirtilen arayüz tanımlı mı diye bakar.

Değiştirgeler

arayüz

Arayüzün ismi.

özdevinimli_yükle

__autoload çağrısı yapılıp yapılmayacağı. Öntanımlı olarak yapılmaz.

Dönen Değerler

Belirtilen arayüz tanımlıysa TRUE, aksi takdirde FALSE döner.

Örnekler

Örnek 1 - interface_exists() örneği

<?php
// Kullanmaya çalışmadan önce arayüz tanımlı mı diye bakalım
if (interface_exists('Arayüzüm')) {
    class 
Sınıfım implements Arayüzüm
    
{
        
// Yöntemler
    
}
}

?>

Ayrıca Bakınız



add a note add a note User Contributed Notes interface_exists
nils dot rocine at gmail dot com 26-Oct-2011 06:35
A little note on namespaces that may be obvious to some, but was not obvious to me.

Although you can make the below statement when the statement is in the same namespace as the interface/class declaration MyInterface...
<?php
$foo
instanceof MyInterface
?>

Making use of the interface_exists, or class_exists functions, you must enter the full namespaced interface name like so (even if the function call is from the same namespace.)
<?php
interface_exists
(__NAMESPACE__ . '\MyInterface', false);
?>
andrey at php dot net 24-Oct-2004 04:40
As far as I remember interface_exists() was added in 5.0.2 . In 5.0.0 and 5.0.1 class_exists() used to return TRUE when asked for a existing interface. Starting 5.0.2 class_exists() doesn't do that anymore.

 
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