That should do what you are trying to do, though I would say it was
better before. You essentially broke the web.
Let your users know that it is a PDF, and let them decide whether to
download or open the file themselves. That should be a browser-side
decision. If your users want to save it, they can right click the link,
just like they do other places on the web.
Cheers,
Cliff
On Tue, 2009-09-22 at 01:19 -0700, luca72 wrote:
> i Have solved using this:
> luca = HttpResponse(open('/home/luca72/webapps/django/myproject/
> disegni/'+nome, 'rb').read(), mimetype='application/octet-stream')
> luca['Content-Disposition'] = 'attachment; filename='+nome
> return luca
>
> Is it ok or i can make it better
>
> Thanks
>
> Luca
>
> On 22 Set, 10:05, luca72 <[email protected]> wrote:
> > So i have try this:
> > file_da_scar = F_d.objects.get(pk=id)
> > nom = file_da_scar.nome_fil
> > return HttpResponse(open('/home/luca72/webapps/django/myproject/
> > disegni/'+nom+'.pdf', 'r').read(), mimetype='application/pdf')
> >
> > And it open the pdf file, but if i need only to download without open
> > it how i
> > can proceed.
> >
> > Thanks
> >
> > Luca
> >
> > On 22 Set, 09:02, luca72 <[email protected]> wrote:
> >
> > > Thanks I use cherry py because i don't know how to download file with
> > > django, can you tell me how to do it with django
> >
> > > regards
> >
> > > Luca
> >
> > > On 14 Set, 12:46, "J. Cliff Dyer" <[email protected]> wrote:
> >
> > > > On Mon, 2009-09-14 at 07:57 -0700, luca72 wrote:
> > > > > Hello i have try with cherrypy but i get this error:
> >
> > > > > def scarico(request, id):
> > > > > from cherrypy.lib.static import serve_file
> > > > > fil_da_scar = F_d.objects.get(pk=id)
> > > > > nome_file = fil_da_scar.nome_fil
> > > > > return serve_file('/home/luca111/Desktop/Luca/Webframework/
> > > > > off_bert/disegni/'+nome_file+'.pdf','application/x-download',
> > > > > 'attachment')
> >
> > > > > Traceback (most recent call last):
> >
> > > > > File "/usr/local/lib/python2.6/site-packages/django/core/servers/
> > > > > basehttp.py", line 278, in run
> > > > > self.result = application(self.environ, self.start_response)
> >
> > > > > File "/usr/local/lib/python2.6/site-packages/django/core/servers/
> > > > > basehttp.py", line 636, in __call__
> > > > > return self.application(environ, start_response)
> >
> > > > > File "/usr/local/lib/python2.6/site-packages/django/core/handlers/
> > > > > wsgi.py", line 245, in __call__
> > > > > response = middleware_method(request, response)
> >
> > > > > File "/usr/local/lib/python2.6/site-packages/django/middleware/
> > > > > common.py", line 83, in process_response
> > > > > if response.status_code == 404:
> >
> > > > > AttributeError: 'generator' object has no attribute 'status_code'
> >
> > > > > What is wrong?
> >
> > > > > Regards
> >
> > > > > Luca
> >
> > > > Well, assuming that scarico is getting called as "middleware_method
> > > > within the code, I'd have to say your problem is that cherrypy's
> > > > serve_file returns a generator object of some sort, rather than a django
> > > > HttpResponse object. This is not surprising, as cherrypy has no reason
> > > > to interoperate properly with django. What you need to do is either
> > > > catch the generator returned by serve_file, and adapt it to an
> > > > HttpResponse.
> >
> > > > Something like.
> >
> > > > def cherrypy_to_HttpResponse(generator):
> > > > #???
> > > > return HttpResponse(*args, **kwargs)
> >
> > > > def scarico(request, id):
> > > > # same as before
> > > > served_file = serve_file('/home/luca111/Desktop/Luca/Webframework/'
> > > > + 'off_bert/disegni/'+nome_file+'.pdf','application/x-download',
> > > > 'attachment')
> > > > return cherrypy_to_HttpResponse(served_file)
> >
> > > > Or figure out a way to do it directly in django.
> >
> > > > I'm not sure exactly what you're trying to do, but if you just return
> > > > the content of the file with an appropriate Content-type set in your
> > > > response headers, I think it will work fine.
> >
> > > > Cheers,
> > > > Cliff
> >
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