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Request #32603 there is no easy way to check if class implements some interface
Submitted: 2005-04-06 08:56 UTC Modified: 2006-04-03 13:46 UTC
From: indeyets at gmail dot com Assigned:
Status: Not a bug Package: Feature/Change Request
PHP Version: 5.0.4 OS:
Private report: No CVE-ID: None
View Developer Edit
 [2005-04-06 08:56 UTC] indeyets at gmail dot com 
Description:
------------
there is no easy way to check if class implements some interface. This
is needed, when, for example, php-application has support for loading
external classes.

external classes have to implement some interface. And check for this
should happen BEFORE object creation. (for example, there might be a
need for some specific constructor syntax).

PHP 5.0 allows to do the following things:
1). $parent = get_parent_class("SomeClassName"). This would be
sufficient, if plugins _extend_ some base class. that's not our case -
wouldn't work for interfaces
2). if ($obj instanceof "SomeInterfaceName") {}. This would work, if we
could create object before the interface check. Wouldn't work for
non-existen objects
3). reflection API. it can do the thing, but overhead (both in code and
in resources) is too big

Patches

Pull Requests

History

AllCommentsChangesGit/SVN commitsRelated reports
 [2005-04-13 00:05 UTC] helly@php.net 
How about: http://php.net/class-implements
 [2005-04-13 07:19 UTC] indeyets at gmail dot com 
class_implements() requires object instantiaiton too. That is a step, which I need to skip.

I need to check if Class implements interface, not Object!
 [2006-04-03 13:46 UTC] tony2001@php.net 
class_implements():
Parameters
class 
An object (class instance) or __a string (class name)__.
 
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