PHP 8.0.3 Released!

# is_a

(PHP 4 >= 4.2.0, PHP 5, PHP 7, PHP 8)

is_aChecks if the object is of this class or has this class as one of its parents

is_a ( mixed $object , string$class_name , bool $allow_string = false ) : bool Checks if the given object is of this class or has this class as one of its parents. ### Parameters object A class name or an object instance. class_name The class name allow_string If this parameter set to false, string class name as object is not allowed. This also prevents from calling autoloader if the class doesn't exist. ### Return Values Returns true if the object is of this class or has this class as one of its parents, false otherwise. ### Examples Example #1 is_a() example <?php // define a class class WidgetFactory { var$oink 'moo';
}

// create a new object
$WF = new WidgetFactory(); if ( is_a($WF'WidgetFactory')) {
echo
"yes, \$WF is still a WidgetFactory\n"; } ?> Example #2 Using the instanceof operator in PHP 5 <?php if ($WF instanceof WidgetFactory) {
echo
'Yes, $WF is a WidgetFactory'; } ?> ### See Also • get_class() - Returns the name of the class of an object • get_parent_class() - Retrieves the parent class name for object or class • is_subclass_of() - Checks if the object has this class as one of its parents or implements it add a note ### User Contributed Notes 7 notes 44 Ronald Locke 4 years ago Please note that you have to fully qualify the class name in the second parameter. A use statement will not resolve namespace dependencies in that is_a() function. <?php namespace foo\bar; class A {}; class B extends A {}; ?> <?php namespace har\var; use foo\bar\A;$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.
33
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'.

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...
11
p dot scheit at zweipol dot net
14 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
15 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($test, "Dog"). This seemed intuitive to me, but did not seem to be documented. dkrupyanskiy[at]gmail 3 years ago Looks like the function signature given in description is wrong. Actually it can take a string as a first parameter in the case if$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{}

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
9 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.
-20
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:
<?php
$object = new \stdClass();$class_name = '\stdClass';

var_dump(is_a($object,$class_name));     // bool(true)
var_dump(is_a($object, '\stdClass')); // bool(true) var_dump($object instanceof $class_name); // bool(true) ?> While this don't: <?php$object
= new \stdClass();
var_dump(\$object instanceof '\stdClass'); // Parse error: syntax error, unexpected ''\stdClass'' (T_CONSTANT_ENCAPSED_STRING)
?>