phpspec - 方法返回对象而不是字符串

Question

我在phpspec中仍然很新鲜,但通常我会在遇到困难时找到解决方案,但这个很难.

我尝试了很多不同的方法,但我还没有找到解决方案.我正在使用Symfony2.

我有一个我想测试的课程:

class MyClass
{

    public function getDataForChildren(MyObject $object)
    {
        foreach ($object->getChildren() as $child) {
            $query = \json_decode($child->getJsonQuery(), true);
            $data = $this->someFetcher->getData($query);
            $child->setData($data);
        }
        return $object;
    }

}

以下是我的spec类:

class MyClassSpec
{

    function let(SomeFetcher $someFetcher)
    {
        $this->beConstructedWith($someFetcher);
    }

    function it_is_initializable()
    {
        $this->shouldHaveType('MyClass');
    }

    function it_should_get_data_for_children_and_return_object(
        MyClass $object,
        MyClass $child, // it means that MyClass has a self-reference to MyClass
        $someFetcher
    )
    {
        $query = '{"id":1}';

        $returnCollection = new ArrayCollection(array($child));

        $object->getChildren()->shouldBeCalled()->willReturn($returnCollection);

        $child->getJsonQuery()->shouldBeCalled()->willReturn($query);

        $someFetcher->getData($query)->shouldBeCalled();

        $this->getDataForChildren($object);
    }

}

运行phpspec后,我收到此错误:

warning: json_decode() expects parameter 1 to be string, object given in

我不知道如何解决这个问题.如果有人有线索,请帮忙.

Answer 1

这是PhpSpec的一个常见绊脚石,声明:

   MyClass $child

表示将使用MyClass的相同接口设置$ child的Collaborator对象.当在SUT(您正在测试的类)中调用child-> getJsonQuery()时,它将返回MethodProphecy,而不是您希望它返回的字符串.

你想说的是你的ArrayCollection 不包含$ child本身(它是一个Collaborator对象),而是包含协作者的真实对象.你这样做:

$returnCollection = new ArrayCollection(array($child->getWrappedObject()));

此外,你不应该在同一个协作者上使用(即多余的)shouldBeCalled()和willReturn(),其中一个就足够了.如果你已经指定了合作者将返回的内容,很明显它将在SUT中被调用.shouldBeCalled()应该用在测试的"assert"部分,以确认Collaborator是用预期的参数调用的,或者是在正确的时间调用的.

您的最终SUT和规格应如下所示:

   class MyClass
   {

        /**
         * @var SomeFetcher
         */
        private $someFetcher;

        public function getDataForChildren(MyObject $object)
        {
            foreach ($object->getChildren() as $child) {
                $query = \json_decode($child->getJsonQuery(), true);
                $data = $this->someFetcher->getData($query);
                $child->setData($data);
            }
            return $object;
        }

        public function getJsonQuery()
        {
        }

        public function setData()
        {
        }

        public function __construct(SomeFetcher $someFetcher)
        {
            $this->someFetcher = $someFetcher;
        }
    }

class MyClassSpec extends ObjectBehavior
{

    function let(SomeFetcher $someFetcher)
    {
        $this->beConstructedWith($someFetcher);
    }

    function it_should_get_data_for_children_and_return_object(
        MyObject $object,
        MyClass $child, // it means that MyClass has a self-reference to MyClass
        SomeFetcher $someFetcher
    )
    {
        $query = '{"id":1}';

        $returnCollection = new ArrayCollection(array($child->getWrappedObject()));

        $object->getChildren()->willReturn($returnCollection);

        $child->getJsonQuery()->willReturn($query);
        $child->setData(Argument::any())->shouldBeCalled();

        $someFetcher->getData(array('id' => 1))->shouldBeCalled();

        $this->getDataForChildren($object);
    }

}

还有,行

$query = \json_decode($child->getJsonQuery(), true);

将在$ query中生成一个关联的数组,即array('id'=> 1)(这是json_encode规定的第二个'true'参数),因此你希望调用$ someFetcher-> getData()后者因此:

$someFetcher->getData(array('id' => 1))->shouldBeCalled();