PHP 7.1.0 Release Candidate 3 Released

is_a

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

is_aVerifica se o objeto é de uma classe ou de sua classe pai

Descrição

bool is_a ( object $object , string $class_name )

Verifica se o dado object é de uma classe ou de sua classe pai.

Nota:

A função is_a() tornou-se obsoleta no PHP 5 pelo operador instanceof.

Parâmetros

object

The tested object

class_name

The class name

Valor Retornado

Returns TRUE if the object is of this class or has this class as one of its parents, FALSE otherwise.

Exemplos

Exemplo #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";
}
?>

Exemplo #2 Using the instanceof operator in PHP 5

<?php
if ($WF instanceof WidgetFactory) {
    echo 
'Yes, $WF is a WidgetFactory';
}
?>

Veja Também

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User Contributed Notes 6 notes

up
15
Aron Budinszky
5 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...
up
11
p dot scheit at zweipol dot net
9 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
up
4
cesoid at yahoo dot com
10 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.
up
2
eitan at mosenkis dot net
4 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.
up
0
Ronald Locke
53 minutes 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.
up
-1
portugal {at} jawira {dot} com
1 year 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)
?>
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