PHPUnit中的MocksvsStubs

我知道存根验证状态,模拟验证行为.

如何在PHPUnit中进行模拟以验证方法的行为？Phpunit没有验证方法(verify()),而且我不知道如何使moks成为PHPUnit.

在文档中,创建存根很好地解释了:

// Create a stub for the SomeClass class.
$stub = $this->createMock(SomeClass::class);

// Configure the stub.
$stub
    ->method('doSomething')
    ->willReturn('foo');

// Calling $stub->doSomething() will now return 'foo'.
$this->assertEquals('foo', $stub->doSomething());

但在这种情况下,我正在验证状态,说回复一个答案.

如何创建模拟和验证行为的示例？

PHPUnit支持两种创建测试双精度的方法.在传统的PHPUnit模拟框架旁边,它现在支持开箱即用的预言.

该createMock方法用于创建三个主要已知的测试双精度.这是你配置对象使它成为虚拟,存根或模拟的方式.

您还可以使用模拟构建器创建测试存根(getMockBuilder返回模拟构建器).这只是做同样事情的另一种方式,它允许您使用流畅的界面调整一些额外的模拟选项(有关更多信息,请参阅文档).

假

虚拟传递,但从未实际调用,或者如果它被调用它响应默认答案(主要是null).它主要用于满足参数列表.

$dummy = $this->createMock(SomeClass::class);

// SUT - System Under Test
$sut->action($dummy);

$stub = $this->createMock(SomeClass::class);
$stub->method('getSomething')
    ->willReturn('foo');

$sut->action($stub);

$mock = $this->createMock(SomeClass::class);
$mock->expects($this->once())
    ->method('doSomething')
    ->with('bar');

$sut->action($mock);

$dummy = $this->prophesize(SomeClass::class);

$sut->action($dummy->reveal());

$stub = $this->prophesize(SomeClass::class);
$stub->getSomething()->willReturn('foo');

$sut->action($stub->reveal());

$mock = $this->prophesize(SomeClass::class);
$mock->doSomething('bar')->shouldBeCalled();

$sut->action($mock->reveal());

$spy = $this->prophesize(SomeClass::class);

// execute the action on system under test
$sut->action($spy->reveal());

// verify expectations after 
$spy->doSomething('bar')->shouldHaveBeenCalled();