Pass something in the request, like ?output=csv

Then look for it in the view and feed the appopriate version.

On Mar 19, 10:35 am, Jesse <[email protected]> wrote:
> My concern is that each view has 87 lines of duplicate code from the
> "GET" data to get the appended list (shown here as: data retrieved to
> create a list .....list.append(publications)).  Anytime I make a
> change in one view I have to remember to make the change in the other
> view.  Since I'm fairly new at Django, I'm thinking there must be a
> more efficient way to write the views?  I'm very pleased with how
> everything is working, so now I am going back to try to clean up the
> code.
>
> Thanks.
>
> On Mar 18, 6:43 pm, Malcolm Tredinnick <[email protected]>
> wrote:
>
> > On Wed, 2009-03-18 at 11:24 -0700, Jesse wrote:
> > > In the view.py I have  (def Bypub) that sends data to a template and
> > > paginates it.  I have a second (def TextFile) that uses the same data
> > > to send to a CSV file.  I would like to combine the two (def Bypub and
> > > def TextFile) together into def Bypub, but I'm not sure how to write
> > > the statements to combine two responses.
>
> > What problem are you trying to solve here? You already have two views
> > where the common stuff is in a separate function (the bit you've marked
> > as retrieving the data in a list) and the distinct stuff is in the
> > separate views. That ideal.
>
> > If the problem is that you want one URL to be able to access both
> > formats, then simplest is to create a third view that works out which of
> > the current two views to call and then passes onto them:
>
> >         def dispatch_view(request,...):
> >            # ... work out which to call somehow...
>
> >            if output_format == 'csv':
> >               return TextFile(request)
> >            else:
> >               return Bypub(request)
>
> > Is that what you're trying to do?
>
> > Regards,
> > Malcolm
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