Question

我正在尝试构建一个查询外部API的类。与端点对应的每种方法都会拨打一个主呼叫＆＃39;负责实际向API发送请求的方法。

例如：

// $this->http is Guzzlehttp\Client 5.3

public function call($httpMethod, $endpoint, array $parameters = [])
{
    $parameters = array_merge($parameters, [
        'headers' => [
            'something' => 'something'
        ]
    ]);

    $request = $this->http->createRequest($httpMethod, $this->baseUrl . $endpoint, $parameters);

    return $this->http->send($request);
}

public function getAll()
{
    return $this->call('GET', 'all');
}

我应该嘲笑什么？我应该在http客户端willBeCalled()和willReturn()方法上使用createRequest()和/或send()吗？

当我模仿send()时，它说：Argument 1 passed to Double\GuzzleHttp\Client\P2::send() must implement interface GuzzleHttp\Message\RequestInterface, null given 而且我不确定如何为此提供假货，因为为该界面创建一个假人需要我在该类上实现30个方法。

现在进行测试：

function it_lists_all_the_things(HttpClient $http)
{
    $this->call('GET', 'all')->willBeCalled();
    $http->createRequest()->willBeCalled();
    $http->send()->willReturn(['foo' => 'bar']);

    $this->getAll()->shouldHaveKeyWithValue('foo', 'bar'); 
}

Answer 1

你应该嘲笑这种行为，如下所示：

public function let(Client $http)
{
    $this->beConstructedWith($http, 'http://someurl.com/');
}

function it_calls_api_correctly(Client $http, Request $request)
{
    $parameters = array_merge([
        'headers' => [
            'something' => 'something'
        ]
    ]);

    $http->createRequest('GET', 'http://someurl.com/all', $parameters)->shouldBeCalled()->willReturn($request);

    $http->send($request)->shouldBeCalled();

    $this->getAll();
}

在phpspec中测试涉及Guzzle

1 个答案: