Python has a very sorry state when it comes to posting something somewhere, even though there is a 2 years old bug for it multipart/form-data encoding - current target is python 2.7 and 3.2.

If the url uses the http: scheme identifier, the optional data argument may be given to specify a POST request (normally the request type is GET). The data argument must be in standard application/x-www-form-urlencoded format; see the urlencode() function below.
urllib — Open arbitrary resources by URL

So, basically it just opens the server connection, and you can compile the body of your post yourself.

import httplib, mimetypes
 
def post_multipart(host, selector, fields, files):
    """
    Post fields and files to an http host as multipart/form-data.
    fields is a sequence of (name, value) elements for regular form fields.
    files is a sequence of (name, filename, value) elements for data to be uploaded as files
    Return the server's response page.
    """
    content_type, body = encode_multipart_formdata(fields, files)
    h = httplib.HTTP(host)
    h.putrequest('POST', selector)
    h.putheader('content-type', content_type)
    h.putheader('content-length', str(len(body)))
    h.endheaders()
    h.send(body)
    errcode, errmsg, headers = h.getreply()
    return h.file.read()
 
def encode_multipart_formdata(fields, files):
    """
    fields is a sequence of (name, value) elements for regular form fields.
    files is a sequence of (name, filename, value) elements for data to be uploaded as files
    Return (content_type, body) ready for httplib.HTTP instance
    """
    BOUNDARY = '----------ThIs_Is_tHe_bouNdaRY_$'
    CRLF = '\r\n'
    L = []
    for (key, value) in fields:
        L.append('--' + BOUNDARY)
        L.append('Content-Disposition: form-data; name="%s"' % key)
        L.append('')
        L.append(value)
    for (key, filename, value) in files:
        L.append('--' + BOUNDARY)
        L.append('Content-Disposition: form-data; name="%s"; filename="%s"' % (key, filename))
        L.append('Content-Type: %s' % get_content_type(filename))
        L.append('')
        L.append(value)
    L.append('--' + BOUNDARY + '--')
    L.append('')
    body = CRLF.join(L)
    content_type = 'multipart/form-data; boundary=%s' % BOUNDARY
    return content_type, body
 
def get_content_type(filename):
    return mimetypes.guess_type(filename)[0] or 'application/octet-stream'

Source: code.activestate.com/recipes/146306

If you do not want to wait for 2.7/3.2 to happen, you have to install pycurl, which is already ported to python3.

Using pycurl has some advantages, for example it does not copy the complete file to post into memory, as it uses a read-callback on the file.

#! /usr/bin/env python
# -*- coding: iso-8859-1 -*-
# vi:ts=4:et
# $Id: file_upload.py,v 1.5 2005/02/13 08:53:13 mfx Exp $
 
import os, sys
import pycurl
 
# Class which holds a file reference and the read callback
class FileReader:
    def __init__(self, fp):
        self.fp = fp
    def read_callback(self, size):
        return self.fp.read(size)
 
# Check commandline arguments
if len(sys.argv) < 3:
    print "Usage: %s <url> <file to upload>" % sys.argv[0]
    raise SystemExit
url = sys.argv[1]
filename = sys.argv[2]
 
if not os.path.exists(filename):
    print "Error: the file '%s' does not exist" % filename
    raise SystemExit
 
# Initialize pycurl
c = pycurl.Curl()
c.setopt(pycurl.URL, url)
c.setopt(pycurl.UPLOAD, 1)
 
# Two versions with the same semantics here, but the filereader version
# is useful when you have to process the data which is read before returning
if 1:
    c.setopt(pycurl.READFUNCTION, FileReader(open(filename, 'rb')).read_callback)
else:
    c.setopt(pycurl.READFUNCTION, open(filename, 'rb').read)
 
# Set size of file to be uploaded.
filesize = os.path.getsize(filename)
c.setopt(pycurl.INFILESIZE, filesize)
 
# Start transfer
print 'Uploading file %s to url %s' % (filename, url)
c.perform()
c.close()

Source: pycurl/examples/file_upload.py