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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