is_a()

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

如果对象属于该类或该类是此对象的父类则返回TRUE

说明

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

如果$object是该类或该类是此对象的父类。

参数

$object

The tested object

$class_name

The class name

$allow_string

If this parameter set toFALSE, string class name as$objectis not allowed. This also prevents from calling autoloader if the class doesn't exist.

返回值

ReturnsTRUEif the object is of this class or has this class as one of its parents,FALSEotherwise.

更新日志

范例

Example #1is_a()例子

<?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";
}
?>

在 PHP 5 中使用instanceof运算符

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

参见

• get_class() 返回对象的类名
• get_parent_class() 返回对象或类的父类名
• is_subclass_of() 如果此对象是该类的子类，则返回 TRUE

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.

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...

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

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));
?>

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.

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.

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)
?>