Dear Ceej
I would like to upload to a form , from a python script. (Client side)
Web2py do no thave upload (as client) rite?
Thanks for reply anyways.
Regards,
Phyo.
On Thu, Nov 13, 2008 at 7:15 PM, ceej <[EMAIL PROTECTED]> wrote:
>
> Phyo,
>
> To get the data from the form that's submitted you can to
> request.vars.file.file.read():
>
> And to write it to uploads you can do something like
>
> _folder='%(folder)suploads/'%{'folder': request.folder}
> main_token=md5.new(str(datetime.datetime.now
> ())).hexdigest()
> if not os.path.exists(_folder):
> os.makedirs('%sfile'%(_folder))
> file = open('%syou_can_make_a_name.jpg'%(_folder), 'wb')
> file.write(request.vars.file.file.read())
> file.close
>
> On Nov 13, 10:33 am, "Phyo Arkar" <[EMAIL PROTECTED]> wrote:
> > Dear All;
> >
> > How can i upload multipart form data files in python (web2py?)
> >
> > I cannot do it with
> >
> > urllib.urlencode
> > urllib2.urlopen(url,data)
> >
> > Files will not be uploaded , just urls will b posted ..(server backend is
> > php)
> >
> > def upload(torrentfile,nfofile,subgenre='Minimal',type=10):
> > nfo_file=file(nfofile)
> > torrent_file=file(torrentfile)
> > data = {'nfo': nfo_file,
> > 'desc': "",
> > 'file': torrent_file,
> > 'type': type,
> > 'subgenre' : subgenre
> > }
> > uldata =urllib.urlencode(data)
> > req = urllib2.Request(upload_url,uldata)
> > resp = urllib2.urlopen(req)
> > result = resp.read()
> > print result
> >
>
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