Please note that you have to fully qualify the class name in the second parameter.
A use statement will not resolve namespace dependencies in that is_a() function.
<?php
namespace foo\bar;
class A {};
class B extends A {};
?>
<?php
namespace har\var;
use foo\bar\A;
$foo = new foo\bar\B();
is_a($foo, 'A'); // returns false;
is_a($foo, 'foo\bar\A'); // returns true;
?>
Just adding that note here because all examples are without namespaces.
is_a
(PHP 4 >= 4.2.0, PHP 5, PHP 7)
is_a — 객체가 이 클래스나 부모 클래스 중 하나인지 확인
설명
bool is_a
( object
$object
, string $class_name
)
주어진 object가 이 클래스나 부모 클래스 중 하나인지
확인합니다.
인수
-
object -
확인할 객체
-
class_name -
클래스명
반환값
객체가 이 클래스나 부모 클래스 중 하나이면 TRUE, 아니면 FALSE를
반환합니다.
변경점
| 버전 | 설명 |
|---|---|
| 5.3.0 |
이 함수는 배제되지 않고, E_STRICT 경고가 발생하지
않습니다.
|
| 5.0.0 |
is_a() 함수는 instanceof 연산자로 인해
PHP 5에서 배제되었습니다. 이 함수를 호출하면
E_STRICT 경고가 발생합니다.
|
예제
Example #1 is_a() 예제
<?php
// 클래스 정의
class WidgetFactory
{
var $oink = 'moo';
}
// 새 객체 생성
$WF = new WidgetFactory();
if (is_a($WF, 'WidgetFactory')) {
echo "yes, \$WF is still a WidgetFactory\n";
}
?>
Example #2 PHP 5에서 instanceof 연산자 사용하기
<?php
if ($WF instanceof WidgetFactory) {
echo 'Yes, $WF is a WidgetFactory';
}
?>
참고
- get_class() - 객체의 클래스명을 반환
- get_parent_class() - 객체나 클래스의 부모 클래스명을 얻습니다
- is_subclass_of() - 이 클래스가 부모 클래스의 자식인지 확인
add a note
User Contributed Notes 7 notes
Ronald Locke ¶
8 years ago
Aron Budinszky ¶
13 years ago
Be careful! Starting in PHP 5.3.7 the behavior of is_a() has changed slightly: when calling is_a() with a first argument that is not an object, __autoload() is triggered!
In practice, this means that calling is_a('23', 'User'); will trigger __autoload() on "23". Previously, the above statement simply returned 'false'.
More info can be found here:
https://bugs.php.net/bug.php?id=55475
Whether this change is considered a bug and whether it will be reverted or kept in future versions is yet to be determined, but nevertheless it is how it is, for now...
p dot scheit at zweipol dot net ¶
17 years ago
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
cesoid at yahoo dot com ¶
19 years ago
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.
dkrupyanskiy[at]gmail ¶
6 years ago
Looks like the function signature given in description is wrong. Actually it can take a string as a first parameter in the case if $allow_string is set to true.
It took some time to find out how the last parameter should be used. Please consider the following example
<?php
class Foo{}
spl_autoload_register(
function($classname){
printf('autoload has been triggered for %s%s', $classname, PHP_EOL);
}
);
var_dump(is_a('UndefinedClassName', Foo::class, true));
?>
eitan at mosenkis dot net ¶
12 years ago
As of PHP 5.3.9, is_a() seems to return false when passed a string for the first argument. Instead, use is_subclass_of() and, if necessary for your purposes, also check if the two arguments are equal, since is_subclass_of('foo', 'foo') will return false, while is_a('foo', 'foo') used to return true.
