我正在使用Python的requests http://docs.python-requests.org/en/latest/index.html我的应用程序的一种方法中的库。该方法的主体如下所示:
def handle_remote_file(url, **kwargs):
response = requests.get(url, ...)
buff = StringIO.StringIO()
buff.write(response.content)
...
return True
我想为该方法编写一些单元测试,但是,我想做的是传递一个假的本地 url,例如:
class RemoteTest(TestCase):
def setUp(self):
self.url = 'file:///tmp/dummy.txt'
def test_handle_remote_file(self):
self.assertTrue(handle_remote_file(self.url))
当我打电话时请求.get有了本地网址,我得到了KeyError以下例外:
requests.get('file:///tmp/dummy.txt')
/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/requests/packages/urllib3/poolmanager.pyc in connection_from_host(self, host, port, scheme)
76
77 # Make a fresh ConnectionPool of the desired type
78 pool_cls = pool_classes_by_scheme[scheme]
79 pool = pool_cls(host, port, **self.connection_pool_kw)
80
KeyError: 'file'
问题是如何将本地 url 传递给请求.get?
PS:上面的例子是我编的。它可能包含很多错误。
正如 @WooParadog 所解释的,请求库不知道如何处理本地文件。虽然,当前版本允许定义传输适配器 http://docs.python-requests.org/en/latest/user/advanced/?highlight=mount#transport-adapters.
因此,您可以简单地定义自己的适配器,它将能够处理本地文件,例如:
from requests_testadapter import Resp
import os
class LocalFileAdapter(requests.adapters.HTTPAdapter):
def build_response_from_file(self, request):
file_path = request.url[7:]
with open(file_path, 'rb') as file:
buff = bytearray(os.path.getsize(file_path))
file.readinto(buff)
resp = Resp(buff)
r = self.build_response(request, resp)
return r
def send(self, request, stream=False, timeout=None,
verify=True, cert=None, proxies=None):
return self.build_response_from_file(request)
requests_session = requests.session()
requests_session.mount('file://', LocalFileAdapter())
requests_session.get('file://<some_local_path>')
我在用着请求测试适配器 https://github.com/ambv/requests-testadapter上面例子中的模块。
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