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 <malc...@pointy-stick.com>
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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