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

目前最简单的方法:

from unittest import TestCase
from unittest.mock import Mock, patch

from .utils import method_foo


class TestFoo(TestCase):

    @patch.object(utils_requests, "post")  # change to desired method here
    def test_foo(self, mock_requests_post):
        # EXPLANATION: mocked 'post' method above will return some built-in mock, 
        # and its method 'json' will return mock 'mock_data',
        # which got argument 'return_value' with our data to be returned
        mock_data = Mock(return_value=[{"id": 1}, {"id": 2}])
        mock_requests_post.return_value.json = mock_data

        method_foo()

        # TODO: asserts here


"""
Example of method that you can test in utils.py
"""
def method_foo():
    response = requests.post("http://example.com")
    records = response.json()
    for record in records:
        print(record.get("id"))
        # do other stuff here

其他回答

解决请求的一个可能的方法是使用库betamax,它记录所有的请求,之后如果你在相同的url中使用相同的参数发出请求,betamax将使用记录的请求,我一直在用它来测试网络爬虫,它节省了我很多时间。

import os

import requests
from betamax import Betamax
from betamax_serializers import pretty_json


WORKERS_DIR = os.path.dirname(os.path.abspath(__file__))
CASSETTES_DIR = os.path.join(WORKERS_DIR, u'resources', u'cassettes')
MATCH_REQUESTS_ON = [u'method', u'uri', u'path', u'query']

Betamax.register_serializer(pretty_json.PrettyJSONSerializer)
with Betamax.configure() as config:
    config.cassette_library_dir = CASSETTES_DIR
    config.default_cassette_options[u'serialize_with'] = u'prettyjson'
    config.default_cassette_options[u'match_requests_on'] = MATCH_REQUESTS_ON
    config.default_cassette_options[u'preserve_exact_body_bytes'] = True


class WorkerCertidaoTRT2:
    session = requests.session()

    def make_request(self, input_json):
        with Betamax(self.session) as vcr:
            vcr.use_cassette(u'google')
            response = session.get('http://www.google.com')

https://betamax.readthedocs.io/en/latest/

这就是模拟请求的方法。Post,将其更改为HTTP方法

@patch.object(requests, 'post')
def your_test_method(self, mockpost):
    mockresponse = Mock()
    mockpost.return_value = mockresponse
    mockresponse.text = 'mock return'

    #call your target method now

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。

我将演示如何通过将真正的请求与返回相同数据的假请求交换来将编程逻辑与实际的外部库分离。在你看来,如果外部api调用,那么这个过程是最好的

import pytest
from unittest.mock import patch
from django.test import RequestFactory

@patch("path(projectname.appname.filename).requests.post")
def test_mock_response(self, mock_get, rf: RequestFactory):
    mock_get.return_value.ok = Mock(ok=True)
    mock_get.return_value.status_code = 400
    mock_get.return_value.json.return_value = {you can define here dummy response}
    request = rf.post("test/", data=self.payload)
    response = view_name_view(request)

    expected_response = {
        "success": False,
        "status": "unsuccessful",
    }

    assert response.data == expected_response
    assert response.status_code == 400

如果使用pytest:

>>> import pytest
>>> import requests

>>> def test_url(requests_mock):
...     requests_mock.get('http://test.com', text='data')
...     assert 'data' == requests.get('http://test.com').text

摘自官方文件