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Urllib2.urlopen Http Error 403

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print(e.read()) ... 404 b' "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">\n\n\nPage Not Found\n ... Is Certificate validation done completely local? Trick or Treat polyglot more hot questions question feed lang-py about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Each handler knows how to open URLs for a particular URL scheme (http, ftp, etc.), or how to handle an aspect of URL opening, for example HTTP redirections or HTTP cookies. http://crimsonskysoftware.com/http-error/urllib2-http-error-401.html

This HOWTO aims to illustrate using urllib, with enough detail about HTTP to help you through. Encante Serum:If some one needs ex... See section 10 of RFC 2616 for a reference on all the HTTP error codes. However, it's the purpose of this tutorial to explain the more complicated cases, concentrating on HTTP.

Python Requests 403 Forbidden

Try spoofing as a browser. This work: import urllib2, sys from bs4 import BeautifulSoup site = "http://youtube.com" page = urllib2.urlopen(site) soup = BeautifulSoup(page) print soup This not work: import urllib2, sys from bs4 import BeautifulSoup site= What to do when majority of the students do not bother to do peer grading assignment? Why does removing Iceweasel nuke GNOME?

Openers use handlers. What are the German equivalents of “First World War”, “World War I”, and “WWI”? e.g. >>> req = urllib.request.Request('http://www.pretend_server.org') >>> try: urllib.request.urlopen(req) ... Raise Httperror(req.full_url, Code, Msg, Hdrs, Fp) Urllib.error.httperror: Http Error 403: Forbidden Is Certificate validation done completely local?

as for the target, google especially is a tough one, kinda hard to scrape, they have implemented many methods to prevent scraping. –andrean Jan 20 '15 at 6:40 | show 4 Python Requests 403 Error One way to do this is to setup our own ProxyHandler, with no proxies defined. How do really talented people in academia think about people who are less capable than them? Number 2¶ from urllib.request import Request, urlopen from urllib.error import URLError req = Request(someurl) try: response = urlopen(req) except URLError as e: if hasattr(e, 'reason'): print('We failed to reach a server.')

Silly mistake :( –FancyDolphin Mar 7 at 0:41 No worries, that would usually be first on my list of things to try. –Padraic Cunningham Mar 7 at 0:45 add Python Urllib Headers And this is how i did it eventually. In that case, it is convenient to use HTTPPasswordMgrWithDefaultRealm. They provide a very nice API, you should use that. –Daniel Roseman Oct 24 '12 at 18:13 Can you give link?

Python Requests 403 Error

This doesn't apply to the original question, of course, but it's still useful to know. –efotinis Jul 28 '13 at 9:19 add a comment| Your Answer draft saved draft discarded Unfortunately a lot of sites still send different versions to different browsers. [3]The user agent for MSIE 6 is ‘Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; .NET CLR 1.1.4322)' [4]For Python Requests 403 Forbidden Sometimes the status code indicates that the server is unable to fulfil the request. Yolk Urllib2.httperror: Http Error 403: Must Access Using Https Instead Of Http info - this returns a dictionary-like object that describes the page fetched, particularly the headers sent by the server.

How strange is it (as an undergrad) to email a professor from another institution about possibly working in their lab? http://crimsonskysoftware.com/http-error/urllib2-http-error-403-forbidden.html Currently for checking login page, it would be fine.But when we go for testing registration page it become tedious. Solutions? Created using Sphinx 1.3.3. Urllib2 User Agent

Not the answer you're looking for? Now, this is not a Python problem, but rather a problem with the webserver itself, perhaps you need to add a username and password to the values list. In python 2.7.8 I have no problem: import urllib url = "https://ipdb.at/ip/66.196.116.112" html = urllib.urlopen(url).read() and everything is fine. have a peek at these guys My question is whether is it the correct way to try or can i write python script in such a way that it can handle it automatically.

Once you have the two side by side, you can progressively work through it until you figure out which header makes the difference - maybe your script isn't sending Host, or Python User-agent NOTE: The page contains Ajax call that creates the table you probably want to parse. geturl - this returns the real URL of the page fetched.

This is probably temporary and should be fixed soon.

  • By default urllib identifies itself as Python-urllib/x.y (where x and y are the major and minor version numbers of the Python release, e.g. Python-urllib/2.5), which may confuse
  • Not the answer you're looking for?
  • Found a bug?

However I want to move to python 3.4 and there I get HTTP error 403 (Forbidden). Not the answer you're looking for? Browse other questions tagged python http web http-status-code-403 or ask your own question. Urllib Vs Urllib2 How do really talented people in academia think about people who are less capable than them?

Then compare with what your script is doing. Why does Wikipedia list an improper pronunciation of Esperanto? What was that alien in Doctor Who that nobody saw? check my blog This is through the ProxyHandler, which is part of the normal handler chain when a proxy setting is detected.

This allows you to specify a default username and password for a URL. python http urllib2 share|improve this question asked Jul 26 '10 at 15:53 Ram Rachum 17.1k41135252 You might want to URL-encode those parentheses. The only way to fix this will be to fool the server that the request is coming from a webbrowser. How to draw a clock-diagram?

http://wolfprojects.altervista.org/changeua.php share|improve this answer answered Jul 26 '10 at 16:03 Eli 80121119 add a comment| up vote 1 down vote Some websites will block access from scripts to avoid 'unnecessary' usage Player claims their wizard character knows everything (from books). Not the answer you're looking for? Below is an example of one way it was solved but isn't working for me.

For example, you can make an FTP request like so: req = urllib.request.Request('ftp://example.com/') In the case of HTTP, there are two extra things that Request objects allow you to do: First, Why cast an A-lister for Groot? Why was Susan treated so unkindly? In its simplest form you create a Request object that specifies the URL you want to fetch.

more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Many thanks. –MaiTiano Mar 6 '12 at 1:52 It's totally ridiculous that they also block HEAD request which are useful e.g. Specifying User-Agent will solve your problem: import urllib.request req = urllib.request.Request(url, headers={'User-Agent': 'Mozilla/5.0'}) html = urllib.request.urlopen(req).read() NOTE Python 2.x urllib version also receives 403 status, but unlike Python 2.x urllib2 and Describing a shrine, just not a Shinto shrine?

The dictionary is reproduced here for convenience # Table mapping response codes to messages; entries have the # form {code: (shortmessage, longmessage)}. AD Domain Controller for the domain could not be contacted What was that alien in Doctor Who that nobody saw? Development of retrosynthesis plan Describing a shrine, just not a Shinto shrine? I don't know and can't imagine why wikipedia does/would do this, but have you tried spoofing your headers?

This specifies the authentication scheme and a ‘realm'. Status code 403 responses are the result of the web server being configured to deny access, for some reason, to the requested resource by the client. How to import urllib.request and urllib.parse: import urllib.request as urlRequest import urllib.parse as urlParse How to make a GET request: url = "http://www.example.net" # open the url x = urlRequest.urlopen(url) #