Hello Micah,
My view is too commplicated. The q and the pathologyid ( I actually
have two other retrievals that also appended with these) are all
appended into the "list". What I need to figure out is how to send
the "list" to it's own dictionary, which I can then retrieve with a
new view and use that view for the pagination. I can see that the
pagination works better with a simple retrieval. Can you tell me how
to create a new object from "list" that can be retrieved in a new
view?
Thanks.
On Mar 11, 10:20 pm, Micah Ransdell <mjr...@gmail.com> wrote:
> Jesse,
>
> The error you are getting is from "q" not being in your post
> QueryDictionary. This happens because when you click next you are not
> rePOSTing the q variable. Either you can put the q variable into an input
> box and POST it again by making your next link a submit button, or you could
> change q to a querystring parameter and use request.GET to retrieve it. This
> makes it easier to pass on to future pages in your search by putting it into
> the querystring for your next link as well.
>
> I have solved this by putting another variable in my next link like this:
>
> {% if page_obj.has_next %}
> <a href="{% url search %}?q={{ request.REQUEST.q }}&page={{
> page_obj.next_page_number }}">next</a>
> {% endif %}
>
> Micah
>
> On Wed, Mar 11, 2009 at 12:51 PM, Jesse <adles...@gmail.com> wrote:
>
> > The following code is for a search on q and attaches tuples from the
> > search on another model. The search results are appended together to
> > create the final data results "list".
>
> > def Bypub(request):
> > query = request.POST['q']
> > if query:
> > qset = (
> > Q(pubtitlestrip__icontains=query) |
> > )
> > pubresults = Publication.objects.filter(qset).distinct() #
> > original
> > else:
> > pubresults = []
> > pathology_id = request.POST['pathology_id'] #original works
> > p = get_object_or_404(Pathology, pk=pathology_id)
> > pub1=Publication.objects.filter(pathpubcombo__pathology=p)
> > for publication in pubresults:
> > if publication not in list:
> > list.append(publication) # the object going to template
>
> > paginator = Paginator(list, 5)
> > try:
> > page = int(request.GET.get('page', '1'))
> > except ValueError:
> > page = 1
> > try:
> > publications = paginator.page(page)
> > except (EmptyPage, InvalidPage):
> > publications = paginator.page(paginator.num_pages)
> > return render_to_response('search/searchresults_pub.html', {
> > "query": query,
> > "pubresults": pubresults,
> > 'pub1': pub1,
> > 'pathology': p,
> > 'list' : list,
> > 'publications' : publications,
> > },
> > context_instance=RequestContext(request)
> > )
>
> > The template:
>
> > {% for publication in publications.object_list %}
> > {{ publication.pubtitle|safe }}
> > {% endfor %}
> > {% else %}
> > No publications found. Please try a different search.
> > {% endif %}
>
> > <div class="pagination">
> > <span class="step-links">
> > {% if publications.has_previous %}
> > <a href="?page=
> > {{ publications.previous_page_number }}">previous</a>
> > {% endif %}
>
> > <span class="current">
> > Page {{ publications.number }} of
> > {{ publications.paginator.num_pages }}.
> > </span>
>
> > {% if publications.has_next %}
> > <a href="?page={{ publications.next_page_number }}">next</
> > a>
> > {% endif %}
> > </span>
> > </div>
>
> > Every thing works just fine until I select the link for the "Next"
> > page. The URL in browser is:
> >http://127.0.0.1:8000/Search/?page=2
> > ERROR:
> > Exception Type: MultiValueDictKeyError
> > Exception Value:
> > Key 'q' not found in <QueryDict: {}>
>
> > I think when I select "Next" it is returning to the def Bypub, but of
> > course the q is not there the second time. How do I preserver the
> > "list" as a dictionary to output to the next page? This might also
> > answer the other problem I'm having with sending the "list" to a csv
> > file. I also need some way to store the data for the csv file.
> > Any help is appreciated. Thanks.
>
>
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