php 检查实例的类是否实现了接口？
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Checking if an instance's class implements an interface?
提问by Wilco
Given a class instance, is it possible to determine if it implements a particular interface? As far as I know, there isn't a built-in function to do this directly. What options do I have (if any)?
给定一个类实例，是否可以确定它是否实现了特定的接口？据我所知，没有内置函数可以直接执行此操作。我有哪些选择（如果有）？
回答by
interface IInterface
{
}
class TheClass implements IInterface
{
}
$cls = new TheClass();
if ($cls instanceof IInterface) {
echo "yes";
}
You can use the "instanceof" operator. To use it, the left operand is a class instance and the right operand is an interface. It returns true if the object implements a particular interface.
您可以使用“instanceof”运算符。要使用它，左操作数是一个类实例，右操作数是一个接口。如果对象实现了特定接口，则返回 true。
回答by Jess Telford
As therefromherepoints out, you can use class_implements(). Just as with Reflection, this allows you to specify the class name as a string and doesn't require an instance of the class:
正如此处指出的那样，您可以使用class_implements(). 与反射一样，这允许您将类名指定为字符串，并且不需要类的实例：
interface IInterface
{
}
class TheClass implements IInterface
{
}
$interfaces = class_implements('TheClass');
if (isset($interfaces['IInterface'])) {
echo "Yes!";
}
class_implements()is part of the SPL extension.
class_implements()是 SPL 扩展的一部分。
See: http://php.net/manual/en/function.class-implements.php
请参阅：http: //php.net/manual/en/function.class-implements.php
Performance Tests
性能测试
Some simple performance tests show the costs of each approach:
一些简单的性能测试显示了每种方法的成本：
Given an instance of an object
给定一个对象的实例
Object construction outside the loop (100,000 iterations) ____________________________________________ | class_implements | Reflection | instanceOf | |------------------|------------|------------| | 140 ms | 290 ms | 35 ms | '--------------------------------------------' Object construction inside the loop (100,000 iterations) ____________________________________________ | class_implements | Reflection | instanceOf | |------------------|------------|------------| | 182 ms | 340 ms | 83 ms | Cheap Constructor | 431 ms | 607 ms | 338 ms | Expensive Constructor '--------------------------------------------'
Given only a class name
只给出一个类名
100,000 iterations ____________________________________________ | class_implements | Reflection | instanceOf | |------------------|------------|------------| | 149 ms | 295 ms | N/A | '--------------------------------------------'
Where the expensive __construct() is:
昂贵的 __construct() 在哪里：
public function __construct() {
$tmp = array(
'foo' => 'bar',
'this' => 'that'
);
$in = in_array('those', $tmp);
}
These tests are based on this simple code.
这些测试基于这个简单的代码。
回答by Bill Karwin
nlaq points out that instanceofcan be used to test if the object is an instance of a class that implements an interface.
nlaq 指出instanceof可以用来测试对象是否是实现接口的类的实例。
But instanceofdoesn't distinguish between a class type and an interface. You don't know if the object is a classthat happens to be called IInterface.
但instanceof不区分类类型和接口。你不知道，如果对象是一个类恰好是叫IInterface。
You can also use the reflection API in PHP to test this more specifically:
您还可以使用 PHP 中的反射 API 来更具体地测试：
$class = new ReflectionClass('TheClass');
if ($class->implementsInterface('IInterface'))
{
print "Yep!\n";
}
回答by d.raev
Just to help future searches is_subclass_ofis also a good variant (for PHP 5.3.7+):
只是为了帮助将来的搜索is_subclass_of也是一个很好的变体（对于 PHP 5.3.7+）：
if (is_subclass_of($my_class_instance, 'ISomeInterfaceName')){
echo 'I can do it!';
}
回答by Starx
You can also do the following
您还可以执行以下操作
public function yourMethod(YourInterface $objectSupposedToBeImplementing) {
//.....
}
It will throw an recoverable error if the $objectSupposedToBeImplementingdoes not implement YourInterfaceInterface.
如果$objectSupposedToBeImplementing没有实现YourInterface接口，它将抛出一个可恢复的错误。
回答by Pilan
Update
更新
The is_afunctionis missing here as alternative.
此处缺少该is_a功能作为替代。
I did some performance tests to check which of the stated ways is the most performant.
我做了一些性能测试来检查哪种所述方式的性能最高。
Results over 100k iterations
超过 10 万次迭代的结果
instanceof [object] took 7.67 ms | + 0% | ..........
is_a [object] took 12.30 ms | + 60% | ................
is_a [class] took 17.43 ms | +127% | ......................
class_implements [object] took 28.37 ms | +270% | ....................................
reflection [class] took 34.17 ms | +346% | ............................................
Added some dots to actually "feel" see the difference.
添加了一些点以实际“感觉”看到差异。
Generated by this: https://3v4l.org/8Cog7
由此生成：https: //3v4l.org/8Cog7
Conclusion
结论
In case you have an objectto check, use instance oflike mentioned in the accepted answer.
如果您有要检查的对象，请使用instance of已接受答案中提到的方法。
In case you have a classto check, use is_a.
如果您有要检查的课程，请使用is_a.
Bonus
奖金
Given the case you want to instantiate a class based on an interface you require it to have, it is more preformant to use is_a. There is only one exception - when the constructor is empty.
考虑到您想要基于您需要的接口实例化一个类的情况，使用is_a. 只有一种例外——构造函数为空时。
Example:
is_a(<className>, <interfaceName>, true);
例子：
is_a(<className>, <interfaceName>, true);
It will return bool. The third parameter "allow_string" allows it to check class names withoutinstantiating the class.
它将返回bool。第三个参数“ allow_string”允许它在不实例化类的情况下检查类名。
