I\'m currently trying to initiate a file upload with urllib2 and the urllib2_file library. Here\'s my code:
import sys
import urllib2_file
import urllib2
UR
First, there's a third way to run Python programs.
From cmd.exe, type python myprogram.py
. You get a nice log. You don't have to type stuff one line at a time.
Second, check the urrlib2 documentation. You'll need to look at urllib, also.
A Request requires a URL and a urlencoded encoded buffer of data.
data should be a buffer in the standard application/x-www-form-urlencoded format. The urllib.urlencode() function takes a mapping or sequence of 2-tuples and returns a string in this format.
You need to encode your data.
If you're using Python 2.5 or newer, urllib2_file
is both unnecessary and unsupported, so check which version you're using (and perhaps upgrade).
If you're using Python 2.3 or 2.4 (the only versions supported by urllib2_file
), try running the sample code and see if you have the same problem. If so, there is likely something wrong either with your Python or urllib2_file
installation.
EDIT:
Also, you don't seem to be using either of urllib2_file
's two supported formats for POST data. Try using one of the following two lines instead:
d = ['uploaded', open(sys.argv[1:])]
## --OR-- ##
d = {'uploaded': open(sys.argv[1:])}
If you're still on Python2.5, what worked for me was to download the code here:
http://peerit.blogspot.com/2007/07/multipartposthandler-doesnt-work-for.html
and save it as MultipartPostHandler.py
then use:
import urllib2, MultipartPostHandler
opener = urllib2.build_opener(MultipartPostHandler.MultipartPostHandler())
opener.open(url, {"file":open(...)})
or if you need cookies:
import urllib2, MultipartPostHandler, cookielib
cj = cookielib.CookieJar()
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cj), MultipartPostHandler.MultipartPostHandler())
opener.open(url, {"file":open(...)})