測試中phpspec一類涉及狂飲

Question

我試圖建立一個查詢外部API的類。與端點相對應的每種方法都會調用負責實際向API發送請求的「主調用」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'); 
} 

什麼我我應該嘲笑？我應該使用的HTTP客戶端的createRequest()和send()方法willBeCalled()和/或willReturn()？

當我嘲笑send()，它說：Argument 1 passed to Double\GuzzleHttp\Client\P2::send() must implement interface GuzzleHttp\Message\RequestInterface, null given ，我不知道如何爲提供假的，因爲在創建該接口的虛擬要求我實施這個類的30種方法。

這裏的測試現在：

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

你應該嘲笑的行爲，這樣的事情：

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(); 
}