我试图使用python模拟包来模拟python请求模块。让我在下面的场景中工作的基本调用是什么?

在views.py中,我有一个函数,它每次都以不同的响应进行各种request .get()调用

def myview(request):
  res1 = requests.get('aurl')
  res2 = request.get('burl')
  res3 = request.get('curl')

在我的测试类中,我想做类似的事情,但不能确定确切的方法调用

步骤1:

# Mock the requests module
# when mockedRequests.get('aurl') is called then return 'a response'
# when mockedRequests.get('burl') is called then return 'b response'
# when mockedRequests.get('curl') is called then return 'c response'

步骤2:

调用我的视图

步骤3:

验证响应包含'a response', 'b response', 'c response'

我如何完成第1步(模拟请求模块)?


当前回答

只是一个有用的提示给那些仍然挣扎,从urllib或urllib2/urllib3转换到请求,并试图模拟一个响应-我在实现我的模拟时得到了一个稍微令人困惑的错误:

与请求。get(path, auth=HTTPBasicAuth('user', 'pass'), verify=False) as url:

AttributeError: __enter__

当然,如果我知道with如何工作(我不知道),我就会知道这是一个残留的、不必要的上下文(来自PEP 343)。在使用请求库时没有必要,因为它在底层为您做了基本相同的事情。只需删除with并使用bare requests.get(…),就万事大吉了。

其他回答

对于那些不想为pytest安装额外库的人,这里有一个例子。我将在这里复制一些扩展,基于上面的例子:

import datetime

import requests


class MockResponse:
    def __init__(self, json_data, status_code):
        self.json_data = json_data
        self.status_code = status_code
        self.elapsed = datetime.timedelta(seconds=1)

    # mock json() method always returns a specific testing dictionary
    def json(self):
        return self.json_data


def test_get_json(monkeypatch):
    # Any arguments may be passed and mock_get() will always return our
    # mocked object, which only has the .json() method.
    def mock_get(*args, **kwargs):
        return MockResponse({'mock_key': 'mock_value'}, 418)

    # apply the monkeypatch for requests.get to mock_get
    monkeypatch.setattr(requests, 'get', mock_get)

    # app.get_json, which contains requests.get, uses the monkeypatch
    response = requests.get('https://fakeurl')
    response_json = response.json()

    assert response_json['mock_key'] == 'mock_value'
    assert response.status_code == 418
    assert response.elapsed.total_seconds() == 1


============================= test session starts ==============================
collecting ... collected 1 item

test_so.py::test_get_json PASSED                                          [100%]

============================== 1 passed in 0.07s ===============================

对于pytest用户,有一个来自https://pypi.org/project/pytest-responsemock/的方便的fixture

例如,模拟GET到http://some.domain,你可以:

def test_me(response_mock):

    with response_mock('GET http://some.domain -> 200 :Nice'):
        response = send_request()
        assert result.ok
        assert result.content == b'Nice'

如果你想模拟一个假响应,另一种方法是简单地实例化一个基本HttpResponse类的实例,如下所示:

from django.http.response import HttpResponseBase

self.fake_response = HttpResponseBase()

我从Johannes Farhenkrug的答案开始,它对我很有效。我需要模拟请求库,因为我的目标是隔离我的应用程序,不测试任何第三方资源。

然后我读了更多关于python的Mock库,我意识到我可以用python Mock类替换MockResponse类,你可能称之为“Test Double”或“Fake”。

这样做的好处是可以访问assert_called_with, call_args等。不需要额外的库。额外的好处,如“可读性”或“更python化”是主观的,所以它们可能对你有影响,也可能没有。

这是我的版本,更新使用python的Mock而不是test double:

import json
import requests
from unittest import mock

# defube stubs
AUTH_TOKEN = '{"prop": "value"}'
LIST_OF_WIDGETS = '{"widgets": ["widget1", "widget2"]}'
PURCHASED_WIDGETS = '{"widgets": ["purchased_widget"]}'


# exception class when an unknown URL is mocked
class MockNotSupported(Exception):
  pass


# factory method that cranks out the Mocks
def mock_requests_factory(response_stub: str, status_code: int = 200):
    return mock.Mock(**{
        'json.return_value': json.loads(response_stub),
        'text.return_value': response_stub,
        'status_code': status_code,
        'ok': status_code == 200
    })


# side effect mock function
def mock_requests_post(*args, **kwargs):
    if args[0].endswith('/api/v1/get_auth_token'):
        return mock_requests_factory(AUTH_TOKEN)
    elif args[0].endswith('/api/v1/get_widgets'):
        return mock_requests_factory(LIST_OF_WIDGETS)
    elif args[0].endswith('/api/v1/purchased_widgets'):
        return mock_requests_factory(PURCHASED_WIDGETS)
    
    raise MockNotSupported


# patch requests.post and run tests
with mock.patch('requests.post') as requests_post_mock:
  requests_post_mock.side_effect = mock_requests_post
  response = requests.post('https://myserver/api/v1/get_widgets')
  assert response.ok is True
  assert response.status_code == 200
  assert 'widgets' in response.json()
  
  # now I can also do this
  requests_post_mock.assert_called_with('https://myserver/api/v1/get_widgets')

Repl。它的链接:

https://repl.it/@abkonsta/Using-unittestMock-for-requestspost#main.py

https://repl.it/@abkonsta/Using-test-double-for-requestspost#main.py

下面是一个带有请求响应类的解决方案。恕我直言,它更干净。

import json
from unittest.mock import patch
from requests.models import Response

def mocked_requests_get(*args, **kwargs):
    response_content = None
    request_url = kwargs.get('url', None)
    if request_url == 'aurl':
        response_content = json.dumps('a response')
    elif request_url == 'burl':
        response_content = json.dumps('b response')
    elif request_url == 'curl':
        response_content = json.dumps('c response')
    response = Response()
    response.status_code = 200
    response._content = str.encode(response_content)
    return response

@mock.patch('requests.get', side_effect=mocked_requests_get)
def test_fetch(self, mock_get):
     response = requests.get(url='aurl')
     assert ...