php PHPUnit、模拟接口和 instanceof
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PHPUnit, mocked interfaces, and instanceof
提问by Bryan M.
Sometimes in my code, I'll check to see if a particular object implements an interface:
有时在我的代码中，我会检查一个特定的对象是否实现了一个接口：
if ($instance instanceof Interface) {};
However, creating mocks of said interface in PHPUnit, I can't seem to pass that test.
但是，在 PHPUnit 中创建所述接口的模拟，我似乎无法通过该测试。
// class name is Mock_Interface_431469d7, does not pass above check
$instance = $this->getMock('Interface');
I understand that having a class named Interface is different from a class implementing Interface, but I'm not sure how to get deal with this.
我知道有一个名为 Interface 的类与实现 Interface 的类不同，但我不确定如何处理这个问题。
Am I forced to mock a concrete class that implements Interface? Wouldn't that defeat the purpose of using an interface for portability?
我是否被迫模拟一个实现接口的具体类？这不会违背使用接口实现可移植性的目的吗？
Thanks
谢谢
回答by Brendon-Van-Heyzen
there is also assertInstanceOf as of 3.5.0
从 3.5.0 开始，还有 assertInstanceOf
Example:
例子：
$this->assertInstanceOf('\Models\User', $this->userService->findById(1));
回答by martinvium
This works for me:
这对我有用：
$mock = $this->getMock('TestInterface');
$this->assertTrue($mock instanceof TestInterface);
Maybe it's a typo or maybe $instance isn't what you think it is?
也许这是一个错字，或者 $instance 不是你认为的那样？
回答by Йося Гисем
Use PhpUnit function assertInstanceOf.
使用 PhpUnit 函数 assertInstanceOf。
Example:
例子：
$this->assertInstanceOf(ResponseInterface::class, $signInResponse);
