Question

如何测试实际工厂方法的静态方法：

public function hireEmployee(HireEmployeeCommand $command)
{
    $username = new EmployeeUsername($command->getUsername());
    $firstName = $command->getFirstName();
    $lastName = $command->getLastName();
    $phoneNumber = new PhoneNumber($command->getPhoneNumber());
    $email = new Email($command->getEmail());
    $role = new EmployeeRole($command->getRole());

    if ($role->isAdmin()) {
        $employee = Employee::hireAdmin($username, $firstName, $lastName, $phoneNumber, $email);
    } else {
        $employee = Employee::hirePollster($username, $firstName, $lastName, $phoneNumber, $email);
    }

    $this->employeeRepository->add($employee);
}

在这里，我无法模拟Employee对象，但我可以为预期的员工模拟EmployeeRepository::add()方法，但我再次检查员工的状态：< / p>

public function it_hires_an_admin()
{
    $this->employeeRepository
        ->add(Argument::that(/** callback for checking state of Employee object */))
        ->shouldBeCalled();

    $this->hireEmployee(
        new HireEmployeeCommand(self::USERNAME, 'John', 'Snow', '123456789', 'john@snow.com', EmployeeRole::ROLE_ADMIN)
    );
}

我意识到我再次模拟了存储库而不是存根。但是在这里，我对员工更感兴趣的将被添加到存储库而不是如何创建它。因此，我应该模拟存储库，但我不应该关心Employee（没有Argument::that()）的状态？看起来合理，但我不能确定创建的员工是否正确。

Answer 1

您并不需要存根或模拟您的实体或价值对象，因为它们在规范中没有表现出来：

public function it_hires_an_admin()
{
    $this->employeeRepository
        ->add(Argument::is(
            Employee::hireAdmin(self::USERNAME, 'John', 'Snow', '123456789', 'john@snow.com')
        ))
        ->shouldBeCalled();

    $this->hireEmployee(
        new HireEmployeeCommand(
            self::USERNAME, 'John', 'Snow', '123456789', 'john@snow.com', EmployeeRole::ROLE_ADMIN
        )
    );
}

使用phpspec在命令处理程序中测试工厂方法

1 个答案: