如何模拟请求和响应?

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

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

2017-02-17 23:02:59

其他回答

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 ...

2020-12-24 11:38:26

以下是对我有效的方法:

import mock
@mock.patch('requests.get', mock.Mock(side_effect = lambda k:{'aurl': 'a response', 'burl' : 'b response'}.get(k, 'unhandled request %s'%k)))

2013-04-02 21:56:26

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

2017-02-17 23:02:59

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

2022-11-11 19:46:11

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/

2018-09-12 17:43:04

如何模拟请求和响应?

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