测试中phpspec一类涉及狂饮

Question

我试图建立一个查询外部API的类。与端点相对应的每种方法都会调用负责实际向API发送请求的“主调用”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'); 
} 

什么我我应该嘲笑？我应该使用的HTTP客户端的createRequest()和send()方法willBeCalled()和/或willReturn()？

当我嘲笑send()，它说：Argument 1 passed to Double\GuzzleHttp\Client\P2::send() must implement interface GuzzleHttp\Message\RequestInterface, null given ，我不知道如何为提供假的，因为在创建该接口的虚拟要求我实施这个类的30种方法。

这里的测试现在：

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

你应该嘲笑的行为，这样的事情：

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(); 
}