Paginate Class-based View in Django

Question

I am trying to paginate my class-based view. Here is how my view looks:

class IssuesByTitleView(ListView):
    context_object_name = "issue_list"

    def issues(request):
        issue_list = Issue.objects.all()
        ###### Commented out does not work ######
        # paginator = Paginator(issue_list, 24)
        # try:
        #    page = int(request.GET.get('page', '1'))
        # except ValueError:
        #   page = 1
        # try:
        #    issues = paginator.page(page)
        # except (EmptyPage, InvalidPage):
        #    issues = paginator.page(paginator.num_pages)

    def get_queryset(self):
        self.title = get_object_or_404(Title, slug=self.kwargs['title_slug'])
        return Issue.objects.filter(title=self.title).order_by('-number')
    def get_context_data(self, **kwargs):
        context = super(IssuesByTitleView, self).get_context_data(**kwargs)
        context['title'] = self.title
        return context

Here is a sample of my models for some context:

class Title(models.Model):
    CATEGORY_CHOICES = (
    ('Ongoing', 'Ongoing'),    
    ('Ongoing - Canceled', 'Ongoing - Canceled'),
    ('Limited Series', 'Limited Series'),
    ('One-shot', 'One-shot'),
    ('Other', 'Other'),
    )    
    title = models.CharField(max_length=64)
    vol = models.IntegerField(blank=True, null=True, max_length=3)
    year = models.CharField(blank=True, null=True, max_length=20, help_text="Ex) 1980 - present, 1980 - 1989.")
    category = models.CharField(max_length=30, choices=CATEGORY_CHOICES)    
    is_current = models.BooleanField(help_text="Check if the title is being published where Emma makes regular appearances.")
    slug = models.SlugField()
    class Meta:
        ordering = ['title']
    def get_absolute_url(self):
        return "/titles/%s" % self.slug        
    def __unicode__(self):

class Issue(models.Model):
    title = models.ForeignKey(Title)
    number = models.CharField(max_length=20, help_text="Do not include the '#'.")
    ...

Of course, by following the Django docs, the pagination system works when the View is defined by something like this: def view(request):

I'm also wondering how I can pull out the next and previous objects.

I would need a link to the "next issue (with the context of the name and issue number)" and then a "previous issue" link. Please note that simply changing the template link with the next or previous number of the issue is not going to work.

So, if anyone can help me out, that would be great.

Answer 1

Just add paginate_by = 20 to you view class.

class IssuesByTitleView(ListView):
    context_object_name = "issue_list"
    paginate_by = 20

    #More stuff here..

Answer 2

Just like Evan Porter has commented, you can make use of the page_obj context variable to access number, paginatior.num_pages, has_next, has_previous. This is what just saved me from the KeyError['page'] after upgrading from Django 1.4.1 to 1.7, object_list to ListView