Python - How to tell you are logged in with urllib2
When you visit the site normally, we can use urllib2.Request as such.
import urllib2, base64
req = urllib2.Request("http://www.facebook.com/")
base64string = base64.encodestring("%s:%s" % ("username", "password")).replace("\n", "")
req.add_header("Authorization", "Basic %s" % base64string)
requested = urllib2.urlopen(req)
But how do we know if you are logged in or not? Because you could just open the wrong authorization url.
+3
source to share
1 answer
Maybe you should look at request.read () to see what the page you just typed says :) Also look at request.info () for headers sent by the server.
And you have to do it in try: ... except:
order to catch mistakes. See Docs.python.org/2/howto/urllib2.html.
FWIW, the modern approach is to use the query module.
EDIT
Here's an excerpt from some code I wrote a couple of years ago.
import urllib2
def post(url, params):
txdata = urllib.urlencode(params)
try:
# create a request object
req = urllib2.Request(url, txdata)
# and open it to return a handle on the url
handle = urllib2.urlopen(req)
except IOError, e:
print >>sys.stderr, 'We failed to open "%s".' % url
if hasattr(e, 'code'):
print >>sys.stderr, 'We failed with error code - %s.' % e.code
elif hasattr(e, 'reason'):
print >>sys.stderr, "The error object has the following 'reason' attribute :"
print >>sys.stderr, e.reason
print >>sys.stderr, "This usually means the server doesn't exist,"
print >>sys.stderr, "is down, or we don't have an internet connection."
#raise SystemExit, 1
raise
else:
print >>sys.stderr, 'Here are the headers of the page :\n%s\n' % handle.info()
true_url = handle.geturl()
print >>sys.stderr, "\nTrue URL = '%s'\n" % true_url
return true_url
Hope this gives you some ideas.
EDIT 2
To handle cookies, just do this before creating the request object:
# build opener with HTTPCookieProcessor
cookie_handler = urllib2.HTTPCookieProcessor()
opener = urllib2.build_opener(cookie_handler)
urllib2.install_opener(opener)
+1
source to share