“Ouvrir le fichier dans l'URL du réseau Python” Réponses codées

Ouvrir le fichier dans l'URL du réseau Python

from requests_testadapter import Resp

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>')