Please note that I neglected to encode data below:
data = { "file" : open(full_path, "rb") }
data = urllib.urlencode(data)Just in case, Miguel On Wed, Dec 9, 2009 at 12:13 PM, Miguel Lopes <[email protected]> wrote: > On Wed, Dec 9, 2009 at 10:26 AM, Miguel Lopes <[email protected]> wrote: >> Hi, >> >> I've been trying to upload a file via a webservice, without any success. >> I'm exposing a serv_upload() action via @service.run. How can I get a >> handle of the file in the serv_upload() action? >> >> I'm using (trying) to use urllib2 to get the file from the server. > > Solved it. My fault for having an outdated version of web2py with a > bad upload bug. > For those eventually interested and for any improvement suggestions, here's > how: > > 1. I used the MultipartPostHandler module by Will Holcomb modified by > Brian Schneider. You can find it here: > http://peerit.blogspot.com/2007/07/multipartposthandler-doesnt-work-for.html > > 2. For the client: > def upload_file(self, full_path): > import MultipartPostHandler, urllib2, cookielib > cookies = cookielib.CookieJar() > opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cookies), > > MultipartPostHandler.MultipartPostHandler) > data = { "file" : open(full_path, "rb") } > try: > response = > opener.open("http://127.0.0.1:8000/myservice/default/call/run/serv_upload/";, > data) > except HTTPError: > log.write(response.read()) > > 3. The easy part :-) the action in web2py: > @service.run > def upload_service(): > import os > infile = request.vars.file > filename = infile.filename > path = os.path.join(request.folder,'private', filename) > f = open(path, 'wb') > f.write(infile.file.read()) > f.close() > > HTH & Txs for any suggestions, > Miguel > -- You received this message because you are subscribed to the Google Groups "web2py-users" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/web2py?hl=en.
