如何訪問url但未在python-response中註冊時發生錯誤？

Question

默認情況下，如果url已註冊但未被訪問，Responses會引發斷言錯誤。如何才能做到相反，即如果一個網址被訪問但未註冊會引發錯誤？

Answer 1

如果要求沒有被註冊的網址，你會得到一個

requests.exceptions.ConnectionError 

例

import responses 
import requests 

@responses.activate 
def test_my_api(): 
    resp = requests.get('https://twitter.com/') 
    print resp 

def run_my_api(): 
    resp = requests.get('https://twitter.com/') 
    print resp 

if __name__ == '__main__': 
    print "Without responses..." 
    run_my_api() 
    print "" 
    print "With responses..." 
    test_my_api() 

輸出

Without responses... 
<Response [200]> 

With responses... 
Traceback (most recent call last): 
    File "example.py", line 18, in <module> 
    test_my_api() 
    File "<string>", line 3, in wrapper 
    File "example.py", line 6, in test_my_api 
    resp = requests.get('https://twitter.com/') 
    File "/usr/local/lib/python2.7/site-packages/requests/api.py", line 70, in get 
    return request('get', url, params=params, **kwargs) 
    File "/usr/local/lib/python2.7/site-packages/requests/api.py", line 56, in request 
    return session.request(method=method, url=url, **kwargs) 
    File "/usr/local/lib/python2.7/site-packages/requests/sessions.py", line 488, in request 
    resp = self.send(prep, **send_kwargs) 
    File "/usr/local/lib/python2.7/site-packages/requests/sessions.py", line 609, in send 
    r = adapter.send(request, **kwargs) 
    File "/usr/local/lib/python2.7/site-packages/responses.py", line 294, in unbound_on_send 
    return self._on_request(adapter, request, *a, **kwargs) 
    File "/usr/local/lib/python2.7/site-packages/responses.py", line 239, in _on_request 
    raise response 
requests.exceptions.ConnectionError: Connection refused: GET https://twitter.com/ 

的<Response [200]>距離undeco打電話給run_my_api。您可以使用assertRaises或等效項檢查異常。