PHP 7.1.2 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

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 classclass 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

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

### User Contributed Notes 6 notes

17
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=55475Whether 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
10 years ago
 At least in PHP 5.1.6 this works as well with Interfaces.<?phpinterface 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
11 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($test, "Dog").This seemed intuitive to me, but did not seem to be documented.  eitan at mosenkis dot net 5 years ago  As of PHP 5.3.9, is_a() seems to return false when passed a string for the first argument. Instead, use is_subclass_of() and, if necessary for your purposes, also check if the two arguments are equal, since is_subclass_of('foo', 'foo') will return false, while is_a('foo', 'foo') used to return true.  Ronald Locke 4 months ago  Please note that you have to fully qualify the class name in the second parameter. A use statement will not resolve namespace dependencies in that is_a() function. <?php namespace foo\bar;class A {};class B extends A {};?><?phpnamespace har\var;use foo\bar\A;$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. 
-2
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)?>