CODEWITHSUNDEEP
pythondjangodjango-formsdjango-templatesdjango-views
Terminal Guy
Terminal Guy

Reputation: 188

How to pass post request data to a separate view? (Django)

I currently have this view which sends a list of genres to a template to populate a dropdown box:

class GenreSearch(generic.ListView):
    template_name = 'games/genresearch.html'
    context_object_name = 'genre_list'

    def get_queryset(self):
        return Genre.objects.order_by("name")

This is the template:

{% if genre_list %}

<div class="btn-group">
  <button class="btn btn-secondary btn-lg dropdown-toggle" type="button" data-toggle="dropdown" aria-haspopup="true" aria-expanded="false">Select Genre</button>
  <div class="dropdown-menu scrollable-menu">
      {% for genre in genre_list %}
        <a class="dropdown-item" href="#">{{ genre.name }}</a>
      {% endfor %}
  </div>
</div>

{% endif %}

What I want to do now is select a genre from this dropdown box, submit it, and use this view to return results based on which genre is submitted:

class GamesPageByGenre(generic.ListView):
    template_name = 'games/allgames.html'
    context_object_name = 'game_list'

    def get_queryset(self):
        #Return games whose genres include the genre id x
        return Game.objects.filter(genres__id=10)

So if the 'action' genre was selected from the dropdown box, submit this, get the ID for the genre action, then replace genres__id=10 with genres__id=genreID

Upvotes: 0

Views: 47

Answers (1)

Vinicius Campanate
Vinicius Campanate

Reputation: 98

Pass the url in dropdown:

<a class="dropdown-item" href="/gamespagebygenre/?genre={{ genre.id }}">{{ genre.name }}</a>

And in the view make like this:

def get_queryset(self):

    genre_id = self.request.GET.get("genre")
    return Game.objects.filter(genres__id=int(genre_id))

Upvotes: 1

Related Questions