CODEWITHSUNDEEP
pythondjangopython-3.x
alcoder
alcoder

Reputation: 457

Pagination with ListView django

I want to create app with search and pagination. Pagination didn't work with ListView.

When I click on the link "next" I am moving from start page http://127.0.0.1:8001/ ---> to the http://127.0.0.1:8001/?city=2 but elements of the list did not change.

And next click to the "next" link did not changes the url ( http://127.0.0.1:8001/?city=2 --> http://127.0.0.1:8001/?city=2).

Could you help me to find error? I think that error in *.html file, but can't find it

My code: models.py

from django.db import models

class City(models.Model):
    name = models.CharField(max_length=255)
    state = models.CharField(max_length=255)

    class Meta:
      verbose_name_plural = "cities"

    def __str__(self):
        return self.name

urls.py

# cities/urls.py
from django.urls import path
from . import views
from .views import HomePageView, SearchResultsView

urlpatterns = [
    path('search/', SearchResultsView.as_view(), name='search_results'),
    path('', HomePageView.as_view(), name='home'),
    path('city/<int:pk>/', views.city_detail, name='city_detail'),
]

views.py

from django.shortcuts import render
from django.views.generic import TemplateView, ListView
from .models import City
from django.db.models import Q
from django.shortcuts import render, get_object_or_404


class HomePageView(ListView):
    model = City
    template_name = 'cities/home.html'
    paginate_by = 3

def city_detail(request, pk):
    city = get_object_or_404(City, pk=pk)
    return render(request, 'cities/city_detail.html', {'city': city})


class SearchResultsView(ListView):
    model = City
    template_name = 'cities/search_results.html'

    def get_queryset(self): # new
        query = self.request.GET.get('q')
        object_list = City.objects.filter(
            Q(name__icontains=query) | Q(state__icontains=query)
        )
        return object_list

home.html

<!-- templates/home.html -->
<h1>HomePage</h1>

<form action="{% url 'search_results' %}" method="get">
  <input name="q" type="text" placeholder="Search...">
</form>

<ul>
  {% for city in object_list %}
    <li>
      <h1><a href="{% url 'city_detail' pk=city.pk %}">{{ city.name }}</a></h1>
    </li>
  {% endfor %}
</ul>


{{page_obj}}

<div class="pagination">
    <span class="page-links">
        {% if page_obj.has_previous %}
            <a href="/?city={{ page_obj.previous_page_number }}">previous</a>
        {% endif %}
            <span class="page-current">
                Page {{ page_obj.number }} of {{ page_obj.paginator.num_pages }}.
            </span>
        {% if page_obj.has_next %}
            <a href="/?city={{ page_obj.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>

Upvotes: 2

Views: 2705

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 477607

The standard name for the page query parameter is 'page' You should either change the name of the queryparameter, or render the template with the ?page= parameter.

Option 1: Changing the page_kwarg

You can change that by altering the page_kwarg attribute [Django-doc]:

class HomePageView(ListView):
    model = City
    template_name = 'cities/home.html'
    paginate_by = 3
    page_kwarg = 'city'

Option 2: Changing the template

It might be more sensical however to simply change the template, such that it uses page as parameter:

<div class="pagination">
    <span class="page-links">
        {% if page_obj.has_previous %}
            <a href="/?page={{ page_obj.previous_page_number }}">previous</a>
        {% endif %}
            <span class="page-current">
                Page {{ page_obj.number }} of {{ page_obj.paginator.num_pages }}.
            </span>
        {% if page_obj.has_next %}
            <a href="/?page={{ page_obj.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>

Upvotes: 6

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