Pagination

Part of the Class-Based Views section of Coddy's Django journey. Lesson 8 of 18.

The jobs page shows every job at once, and a thousand would not fit. Pagination splits a list into pages with links between them. A ListView does the splitting with one attribute, paginate_by, the rows per page.

With paginate_by = 20 on a recipe list, /recipes/ shows the first 20, /recipes/?page=2 the next 20, and a page past the end answers 404. The template's list holds only this page's rows, and the template also gets is_paginated and page_obj, the current page.

The links are yours to write. page_obj knows number, paginator.num_pages, has_next, next_page_number, has_previous and previous_page_number:

{% if page_obj.has_next %}
  <a href="?page={{ page_obj.next_page_number }}">More recipes</a>
{% endif %}

Keep the test around every link: on the last page, next_page_number raises an error that crashes the page.

Challenge

Easy

Show the jobs 5 per page:

In JobListView, 5 rows per page.
In job_list.html, under the list, a <nav class="pages"> with a Previous link, the text Page 1 of 3 (real numbers) and a Next link. Each link shows only when its page exists.

Press Run, open /jobs/ and press Next.

Try it yourself

from django.views.generic import DetailView, ListView, TemplateView

from .models import Job


class HomeView(TemplateView):
    template_name = "jobs/home.html"

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context["count"] = Job.objects.count()
        context["latest"] = Job.objects.order_by("-posted").first()
        return context


class AboutView(TemplateView):
    template_name = "jobs/about.html"


class JobListView(ListView):
    model = Job
    context_object_name = "jobs"
    ordering = ["-posted"]

    def get_queryset(self):
        jobs = super().get_queryset()
        city = self.request.GET.get("city")
        if city:
            jobs = jobs.filter(city=city)
        return jobs


class JobDetailView(DetailView):
    model = Job