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

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

2021-10-25 10:48:47

其他回答

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

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

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'

2021-04-16 02:39:01

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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