(PHP 4 >= 4.2.0, PHP 5, PHP 7)
is_a — 如果对象属于该类或该类是此对象的父类则返回 TRUE
$object
, string $class_name
[, bool $allow_string = FALSE
] )
如果 object 是该类或该类是此对象的父类。
objectThe tested object
class_nameThe 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.
Returns TRUE if the object is of this class or has this class as one of
its parents, FALSE otherwise.
| 版本 | 说明 |
|---|---|
| 5.3.9 |
Added allow_string parameter
|
| 5.3.0 |
This function is no longer deprecated, and will therefore
no longer throw E_STRICT warnings.
|
| 5.0.0 |
This function became deprecated in favour of the
instanceof
operator. Calling this function will result in an
E_STRICT warning.
|
Example #1 is_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";
}
?>
Example #2 在 PHP 5 中使用 instanceof 运算符
<?php
if ($WF instanceof WidgetFactory) {
echo 'Yes, $WF is a WidgetFactory';
}
?>
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.
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)
?>
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.
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
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.