Iterate through a list and try log in to a website with urllib and re
roy at panix.com
Mon Mar 3 14:47:16 CET 2014
In article <fc7c1c7d-7811-4e58-a999-1f2e9b297cd3 at googlegroups.com>,
Marcus <marcuscvj at gmail.com> wrote:
> I'm trying to use urllib and urllib2 to open an url + login_data in a for
Step 1: Ignore all that crap and get http://www.python-requests.org/
> How can I display when successfully logged in and how to show when the
> login is denied?
> I've tried use this:
> html_content = urllib2.urlopen(url).read()
> re.findall('ERROR: The password you entered for the username USER is
> incorrect.', html_content)
In the ideal case, whatever you're talking to will return a success or
failure indication in the HTTP status code.
Lacking that, it will at least return something intended to be parsed
Lacking that (and, unfortunately, common), you're reduced to
screen-scraping. But, if you've got to do that, at least use a tool
like lxml or BeautifulSoup to parse the HTML.
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