interface_exists
At least in PHP 5.1.6 this works as well with Interfaces.
<?php
interface test {
public function A();
}
class TestImplementor implements test {
public function A () {
print "A";
}
}
$testImpl = new TestImplementor();
var_dump(is_a($testImpl,'test'));
?>
will return true
is_a
(PHP 4 >= 4.2.0, PHP 5)
is_a — 객체가 이 클래스에 속해있거나 부모 클래스의 하나가 이 클래스이면, TRUE를 반환한다
설명
bool is_a
( object $object
, string $class_name
)
이 함수는 특정 객체가 이 클래스에 속해있거나 부모 클래스의 하나가 이 클래스이면, TRUE를 반환하고, 그렇지 않으면, FALSE를 반환한다.
add a note
User Contributed Notesis_a
p dot scheit at zweipol dot net
16-Jan-2007 01:44
16-Jan-2007 01:44
martin dunisch
13-Feb-2006 12:02
13-Feb-2006 12:02
Workaround for older PHP-Versions:
function is_a($anObject, $aClass) {
return get_class($anObject) == strtolower($aClass)
or is_subclass_of($anObject, $aClass);
}
dead dot screamer at seznam dot cz
06-Feb-2006 11:44
06-Feb-2006 11:44
Why I test if class `A` inherit class `B` or implements interface `C` before I create class `A`?
<?
//That isn't work:
//1. function is_A()
if(is_A('A','B'))$a=new A;
if(is_A('A','C'))$a=new A;
//2. operator instanceOf
if(A instanceOf B)$a=new A;
if(A instanceOf C)$a=new A;
?>
zabmilenko at hotmail dot com
07-Oct-2005 05:18
07-Oct-2005 05:18
Lazy Instantiation using is_a() and php5
<?php
class ObjectA
{
public function print_line($text)
{
print $text . "\n";
}
}
class ObjectB
{
public function ObjectA()
{
static $objecta;
if (!is_a($objecta, 'ObjectA'))
{
$objecta = new ObjectA;
}
return $objecta;
}
}
$obj = new ObjectB;
$obj->ObjectA()->print_line('testing, 1 2 3');
?>
In the above example, ObjectA is not instantiated until needed by ObjectB. Then ObjectB can continually use it's creation as needed without reinstantiating it.
There are other ways, but I like this one :-)
cesoid at yahoo dot com
05-Oct-2005 07:01
05-Oct-2005 07:01
is_a returns TRUE for instances of children of the class.
For example:
class Animal
{}
class Dog extends Animal
{}
$test = new Dog();
In this example is_a($test, "Animal") would evaluate to TRUE as well as is_a($test, "Dog").
This seemed intuitive to me, but did not seem to be documented.