Thanks Zak... I figured that out, but it's always good to know I
figured something out right.
On Jan 25, 4:35 pm, Zak Johnson <[EMAIL PROTECTED]> wrote:
> [EMAIL PROTECTED] wrote:
> > It looks like it's because (as you said), I'm returning a string, in
> > the exception... so how do I do that?
>
> > def add_song(request):
> > try:
> > f = open('/path/to/songs/'+
> > request.FILES['song_file']['filename'], 'wb') # wb = write binary
> > f.write(request.FILES['song_file']['content'])
> > f.close()
> > except:
> > return HttpResponse("Are you sure you selected a song to
> > upload?", mimetype="text/html")
> > song = Song(artist_id=request.user.id, song_file=
> > "songs/"+request.FILES['song_file']['filename'],song_title=
> > request.POST['song_title'],pub_date=datetime.now(),
> > description=request.POST['description'])
> > song.save()You're correctly returning an HttpResponse object when an
> > exception is
> raised; the problem is that you're not returning anything at the end of
> your function, following the `song.save()`. You should append a line
> like:
>
> return HttpResponseRedirect(some_url)
>
> "some_url" should be the URL for an "upload successful" page.
>
> -Zak
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