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 My fault for having an outdated version of web2py with a bad upload bug.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.
