PhPunit 避免 mock 的构造函数参数

有什么方法可以避免 phPunit 不得不为模拟对象调用构造函数?否则,我将需要一个模拟对象作为构造函数参数,另一个等。API 似乎是这样的:

getMock($className, $methods = array(), array $arguments = array(),
$mockClassName = '', $callOriginalConstructor = TRUE,
$callOriginalClone = TRUE, $callAutoload = TRUE)

我没有让它工作,它仍然抱怨构造函数参数,即使 $callOriginalConstructor设置为 false。

我在构造函数中只有一个对象,它是一个依赖注入。所以我觉得我没有设计上的问题。

71205 次浏览

Perhaps you need to create a stub to pass in as the constructor argument. Then you can break that chain of mock objects.

PHPUnit is designed to call the constructor on mocked objects; to prevent this you should either:

  1. Inject a mock object as a dependency into the object you're having trouble mocking
  2. Create a test class that extends the class you're trying to call that doesn't call the parent constructor

Here you go:

    // Get a Mock Soap Client object to work with.
$classToMock = 'SoapClient';
$methodsToMock = array('__getFunctions');
$mockConstructorParams = array('fake wsdl url', array());
$mockClassName = 'MyMockSoapClient';
$callMockConstructor = false;
$mockSoapClient = $this->getMock($classToMock,
$methodsToMock,
$mockConstructorParams,
$mockClassName,
$callMockConstructor);

You can use getMockBuilder instead of just getMock:

$mock = $this->getMockBuilder('class_name')
->disableOriginalConstructor()
->getMock();

See the section on "Test Doubles" in PHPUnit's documentation for details.

Although you can do this, it's much better to not need to. You can refactor your code so instead of a concrete class (with a constructor) needing to be injected, you only depend upon an interface. This means you can mock or stub the interface without having to tell PHPUnit to modify the constructor behaviour.

As an addendum, I wanted to attach expects() calls to my mocked object and then call the constructor. In PHPUnit 3.7.14, the object that is returned when you call disableOriginalConstructor() is literally an object.

// Use a trick to create a new object of a class
// without invoking its constructor.
$object = unserialize(
sprintf('O:%d:"%s":0:{}', strlen($className), $className)

Unfortunately, in PHP 5.4 there is a new option which they aren't using:

ReflectionClass::newInstanceWithoutConstructor

Since this wasn't available, I had to manually reflect the class and then invoke the constructor.

$mock = $this->getMockBuilder('class_name')
->disableOriginalConstructor()
->getMock();


$mock->expect($this->once())
->method('functionCallFromConstructor')
->with($this->equalTo('someValue'));


$reflectedClass = new ReflectionClass('class_name');
$constructor = $reflectedClass->getConstructor();
$constructor->invoke($mock);

Note, if functionCallFromConstruct is protected, you specifically have to use setMethods() so that the protected method is mocked. Example:

    $mock->setMethods(array('functionCallFromConstructor'));

setMethods() must be called before the expect() call. Personally, I chain this after disableOriginalConstructor() but before getMock().

Alternatively you could add a parameter to getMock to prevent the calling of the default constructor.

$mock = $this->getMock(class_name, methods = array(), args = array(),
mockClassName = '', callOriginalConstructor = FALSE);

Still, I think the answer of dave1010 looks nicer, this is just for the sake of completeness.

This question is a little old, but for new visitors, you can do it using the createMock method (previously called createTestDouble and introduced in v5.4.0).

$mock = $this->createMock($className);

As you can see in the code below extracted from the PHPUnit\Framework\TestCase class (in phpunit/src/framework/TestCase.php), it will basically create a mock object without calling the original constructor.

/** PHPUnit\Framework\TestCase::createMock method */
protected function createMock(string $originalClassName): MockObject
{
return $this->getMockBuilder($originalClassName)
->disableOriginalConstructor()
->disableOriginalClone()
->disableArgumentCloning()
->disallowMockingUnknownTypes()
->getMock();
}