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()

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/

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 ...

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

您可以使用request -mock代替吗?

假设你的myview函数接受一个请求。Session对象，用它发出请求，并对输出做一些事情:

# mypackage.py
def myview(session):
    res1 = session.get("http://aurl")
    res2 = session.get("http://burl")
    res3 = session.get("http://curl")
    return f"{res1.text}, {res2.text}, {res3.text}"

# test_myview.py
from mypackage import myview
import requests

def test_myview(requests_mock):
    # set up requests
    a_req = requests_mock.get("http://aurl", text="a response")
    b_req = requests_mock.get("http://burl", text="b response")
    c_req = requests_mock.get("http://curl", text="c response")

    # test myview behaviour
    session = requests.Session()
    assert myview(session) == "a response, b response, c response"

    # check that requests weren't called repeatedly
    assert a_req.called_once
    assert b_req.called_once
    assert c_req.called_once
    assert requests_mock.call_count == 3

你也可以在Pytest之外的框架中使用requests_mock——文档非常棒。