Thanks for the answers,
@Nick, but the issue is if add the hidden value buscar, I must add all
the parameters used in the search form (quite long), I'm looking for a
lazy way doing it.
@Rodrigo, same issue, if there is no search parameters, I can't apply
my filters to the model, so I will not be able to retrieve the "page
2" related items, isn't it?
Thank you very much for yours answers.
Regards
Bvcelari
On Nov 26, 1:17 pm, "Rodrigo \"OreiA\" Chacon" <rocha...@gmail.com>
wrote:
> The most secure way to retrieve a parameter is to use the get method
> of the QueryDict.
>
> So, instead of:
> request.GET['buscar']
> do:
> request.GET.get('buscar', 'default_value')
>
> Or check if the key exists in the QueryDict instance before trying to
> retrieve it, with:
> if 'buscar' in request.GET:
> buscar = request.GET['buscar']
>
> ;-)
>
> -- Rodrigo Chacon
>
> On Nov 25, 12:42 pm, Nick Arnett <nick.arn...@gmail.com> wrote:
>
> > On Thu, Nov 25, 2010 at 6:35 AM, bvcelari <bvcel...@gmail.com> wrote:
> > > Hy,
> > > I'm trying to deploy my first django app, and I'm trying to use
> > > pagination , I 'm using Django 1.2
> > > I'm using a simple view with form, and with the request show the
> > > results paginated. when I try to go to next page
> > > the message appears:
> > > "Key 'buscar' not found in <QueryDict: {u'page': [u'2']}>"
> > > buscar is a hidden value used for check if the request comes from the
> > > search form.
> > > I think this message comes from the "request" URL is not properly
> > > generated in fact my link targets to
> > > "http://localhost:8000/search/?page=2";
> > > instead of something like this:
> > > "http://localhost:8000/search/?option_value1=7&option_value2=3&page=2";
> > > There is any way to maintain the searchred url and indicate wich is
> > > the next page?
>
> > You need to pass that variable to the template, then add it to the page it
> > as a hidden input.
>
> > For example:
>
> > <input type="hidden" name="buscar" value="{{ buscar }}"></input>
>
> > Nick
>
>
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