In spirit of Dont Repeat Yourself And to avoid Copy Pasting :
os.popen('curl -F "[EMAIL PROTECTED]" -F "[EMAIL PROTECTED]" -F "type=%s" -F
"subgenre=%s" "
http://127.0.0.1:81/ultest/uploader.php";' %
(torrentfile,nfofile,subgenre,type) )
This does the best trick :D
On Thu, Nov 13, 2008 at 9:05 PM, mdipierro <[EMAIL PROTECTED]> wrote:
>
> makes sense. Let us know if the script above does the trick.
>
> Massimo
>
> On Nov 13, 2:49 pm, "Phyo Arkar" <[EMAIL PROTECTED]> wrote:
> > Acutally , I am building a file upload script :D Which automaticlaly
> files
> > to a mirrored server via http.
> >
> > On Thu, Nov 13, 2008 at 7:56 PM, Yarko T <[EMAIL PROTECTED]> wrote:
> > > Phyo is trying to build a bot! ;-)
> >
> > > On Thu, Nov 13, 2008 at 1:45 PM, Phyo Arkar <[EMAIL PROTECTED]
> >wrote:
> >
> > >> 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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