Proxy authentication in common libraries
Hi I am behind a proxy server that needs proxy authentication. There are a lot of libraries that come without proxy support. The function below, which is part of the python-twitter library does HTTP Authentication, and I can't figure out how to do this with a ProxyBasicAuthHandler object. I'm pasting the function here. Can someone tell me how this code can be rewritten to do proxy authentication as well as http authentication? def _GetOpener(self, url, username=None, password=None): if username and password: self._AddAuthorizationHeader(username, password) handler = self._urllib.HTTPBasicAuthHandler() (scheme, netloc, path, params, query, fragment) = urlparse.urlparse(url) handler.add_password(Api._API_REALM, netloc, username, password) opener = self._urllib.build_opener(handler) else: opener = self._urllib.build_opener() opener.addheaders = self._request_headers.items() return opener -- http://mail.python.org/mailman/listinfo/python-list
Re: Odd math related issue.
On Jul 21, 5:30 pm, Fredrik Lundh <[EMAIL PROTECTED]> wrote: > Alexandru Palade wrote: > > However, you should be carefully because using an %i modifier for a > > what-should-be a float value truncates the value in a way you may not > > expect. > > > What I mean is that if you have sent 2 out of 3 bytes, the math will be > > 200/3 which with the %i modifier will print 66, rather than 66.6 (or at > > least 67 which is closer - have a look at the round() function). > > My suggested workaround doesn't use floats. As for rounding, that's > more of a usability issue -- seeing the download process getting stuck > at 100% can be rather frustrating for the poor user. Better truncate > towards zero. > > > Another thing, you could have just added a dot after the constant in > > order to promote the expression to be evaluated as float. As in > > percentage = bytes_transferred / /self/.__sessions[path].total_bytes > > * 100. > > (notice the last dot) > > Did you try that? > > >>> 2 / 3 * 100. > 0.0 > > You can fix this with parentheses, but I usually recommend an explicit > "cast" instead, to make it obvious what you're doing: > > result = float(a) / b > > Use 2.0 / 3 * 100 to solve this. Why make things look bigger? -- http://mail.python.org/mailman/listinfo/python-list
Re: Odd math related issue.
On Jul 21, 5:30 pm, Fredrik Lundh <[EMAIL PROTECTED]> wrote: > Alexandru Palade wrote: > > However, you should be carefully because using an %i modifier for a > > what-should-be a float value truncates the value in a way you may not > > expect. > > > What I mean is that if you have sent 2 out of 3 bytes, the math will be > > 200/3 which with the %i modifier will print 66, rather than 66.6 (or at > > least 67 which is closer - have a look at the round() function). > > My suggested workaround doesn't use floats. As for rounding, that's > more of a usability issue -- seeing the download process getting stuck > at 100% can be rather frustrating for the poor user. Better truncate > towards zero. > > > Another thing, you could have just added a dot after the constant in > > order to promote the expression to be evaluated as float. As in > > percentage = bytes_transferred / /self/.__sessions[path].total_bytes > > * 100. > > (notice the last dot) > > Did you try that? > > >>> 2 / 3 * 100. > 0.0 > > You can fix this with parentheses, but I usually recommend an explicit > "cast" instead, to make it obvious what you're doing: > > result = float(a) / b > > try 2.0 / 3 * 100 that would solve all trouble -- http://mail.python.org/mailman/listinfo/python-list
