PHP Conference China 2020


(PHP 4 >= 4.2.0, PHP 5, PHP 7)

is_aVérifie si l'objet est une instance d'une classe donnée ou a cette classe parmi ses parents


is_a ( mixed $object , string $class_name [, bool $allow_string = FALSE ] ) : bool

Vérifie si l'objet object est une instance d'une classe donnée ou a cette classe comme parent.

Liste de paramètres


Un nom de classe ou une instance d'un objet.


Le nom de la classe


Si ce paramètre vaut FALSE, le nom de la classe sous forme de chaîne de caractères dans le paramètre object n'est pas autorisé. Ceci permet d'éviter l'appel de l'autoloader si la classe n'existe pas.

Valeurs de retour

Retourne TRUE si l'objet est une instance de cette classe ou a cette classe comme parent, FALSE sinon.


Version Description
5.3.9 Ajout du paramètre allow_string.
5.3.0 Cette fonction n'est plus obsolète, et ne lancera donc plus d'alerte de type E_STRICT.
5.0.0 Cette fonction devient obsolète en faveur de l'opérateur instanceof. L'appel à cette fonction émettra une alerte de niveau E_STRICT.


Exemple #1 Exemple avec is_a()

// Définit une classe
class WidgetFactory
$oink 'moo';

// Crée un nouvel objet
$WF = new WidgetFactory();

if (
is_a($WF'WidgetFactory')) {
"oui, \$WF est toujours un objet WidgetFactory\n";

Exemple #2 Utilisation de l'opérateur instanceof avec PHP 5

if ($WF instanceof WidgetFactory) {
'Oui, $WF est un WidgetFactory';

Voir aussi

add a note add a note

User Contributed Notes 7 notes

Ronald Locke
3 years ago
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.

namespace foo\bar;

A {};
B extends A {};

namespace har\var;

$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.
Aron Budinszky
9 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:

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
13 years ago
At least in PHP 5.1.6 this works as well with Interfaces.

interface test {
  public function

TestImplementor implements test {
  public function
A () {

$testImpl = new TestImplementor();


will return true
2 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


class Foo{}

printf('autoload has been triggered for %s%s', $classname, PHP_EOL);

var_dump(is_a('UndefinedClassName', Foo::class, true));

cesoid at yahoo dot com
14 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.
eitan at mosenkis dot net
8 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.
portugal {at} jawira {dot} com
5 years ago
I just want to point out that you can replace "is_a()" function with the "instanceof" operator, BUT you must use a variable to pass the class name string.

This will work:
= new \stdClass();
$class_name = '\stdClass';

var_dump(is_a($object, $class_name));     // bool(true)
var_dump(is_a($object, '\stdClass'));     // bool(true)
var_dump($object instanceof $class_name); // bool(true)

While this don't:
= new \stdClass();
var_dump($object instanceof '\stdClass'); // Parse error: syntax error, unexpected ''\stdClass'' (T_CONSTANT_ENCAPSED_STRING)
To Top