Python urllib2 Basic Auth Problem

Question

Update: based on Lee’s comment I decided to condense my code to a really simple script and run it from the command line:

import urllib2
import sys

username = sys.argv[1]
password = sys.argv[2]
url = sys.argv[3]
print("calling %s with %s:%sn" % (url, username, password))

passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, url, username, password)
urllib2.install_opener(urllib2.build_opener(urllib2.HTTPBasicAuthHandler(passman)))

req = urllib2.Request(url)
f = urllib2.urlopen(req)
data = f.read()
print(data)

Unfortunately it still won’t generate the Authorization header (per Wireshark) 🙁

I’m having a problem sending basic AUTH over urllib2. I took a look at this article, and followed the example. My code:

passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, "api.foursquare.com", username, password)
urllib2.install_opener(urllib2.build_opener(urllib2.HTTPBasicAuthHandler(passman)))

req = urllib2.Request("http://api.foursquare.com/v1/user")    
f = urllib2.urlopen(req)
data = f.read()

I’m seeing the following on the Wire via wireshark:

GET /v1/user HTTP/1.1
Host: api.foursquare.com
Connection: close
Accept-Encoding: gzip
User-Agent: Python-urllib/2.5

You can see the Authorization is not sent, vs. when I send a request via curl: curl -u user:password http://api.foursquare.com/v1/user

GET /v1/user HTTP/1.1
Authorization: Basic =SNIP=
User-Agent: curl/7.19.4 (universal-apple-darwin10.0) libcurl/7.19.4 OpenSSL/0.9.8k zlib/1.2.3
Host: api.foursquare.com
Accept: */*

For some reason my code seems to not send the authentication – anyone see what I’m missing?

thanks

-simon

Asked By: Simon

||

Source

Answer 1

The second parameter must be a URI, not a domain name. i.e.

passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, "http://api.foursquare.com/", username, password)

Answered By: Lee

Answer 2

The problem could be that the Python libraries, per HTTP-Standard, first send an unauthenticated request, and then only if it’s answered with a 401 retry, are the correct credentials sent. If the Foursquare servers don’t do “totally standard authentication” then the libraries won’t work.

Try using headers to do authentication:

import urllib2, base64

request = urllib2.Request("http://api.foursquare.com/v1/user")
base64string = base64.b64encode('%s:%s' % (username, password))
request.add_header("Authorization", "Basic %s" % base64string)   
result = urllib2.urlopen(request)

Had the same problem as you and found the solution from this thread: http://forums.shopify.com/categories/9/posts/27662

Answered By: yayitswei

Answer 3

(copy-paste/adapted from https://stackoverflow.com/a/24048772/1733117).

First you can subclass urllib2.BaseHandler or urllib2.HTTPBasicAuthHandler, and implement http_request so that each request has the appropriate Authorization header.

import urllib2
import base64

class PreemptiveBasicAuthHandler(urllib2.HTTPBasicAuthHandler):
    '''Preemptive basic auth.

    Instead of waiting for a 403 to then retry with the credentials,
    send the credentials if the url is handled by the password manager.
    Note: please use realm=None when calling add_password.'''
    def http_request(self, req):
        url = req.get_full_url()
        realm = None
        # this is very similar to the code from retry_http_basic_auth()
        # but returns a request object.
        user, pw = self.passwd.find_user_password(realm, url)
        if pw:
            raw = "%s:%s" % (user, pw)
            auth = 'Basic %s' % base64.b64encode(raw).strip()
            req.add_unredirected_header(self.auth_header, auth)
        return req

    https_request = http_request

Then if you are lazy like me, install the handler globally

api_url = "http://api.foursquare.com/"
api_username = "johndoe"
api_password = "some-cryptic-value"

auth_handler = PreemptiveBasicAuthHandler()
auth_handler.add_password(
    realm=None, # default realm.
    uri=api_url,
    user=api_username,
    passwd=api_password)
opener = urllib2.build_opener(auth_handler)
urllib2.install_opener(opener)

Answered By: dnozay

Answer 4

I would suggest that the current solution is to use my package urllib2_prior_auth which solves this pretty nicely (I work on inclusion to the standard lib.

Answered By: mcepl

Answer 5

Here’s what I’m using to deal with a similar problem I encountered while trying to access MailChimp’s API. This does the same thing, just formatted nicer.

import urllib2
import base64

chimpConfig = {
    "headers" : {
    "Content-Type": "application/json",
    "Authorization": "Basic " + base64.encodestring("hayden:MYSECRETAPIKEY").replace('n', '')
    },
    "url": 'https://us12.api.mailchimp.com/3.0/'}

#perform authentication
datas = None
request = urllib2.Request(chimpConfig["url"], datas, chimpConfig["headers"])
result = urllib2.urlopen(request)

Answered By: Hayden Shelton

Python urllib2 Basic Auth Problem

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