Question

我正在尝试对基于类的视图进行分页。以下是我的观点：

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

以下是我的模型的一些示例：

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 '#'.")
    ...

当然，通过遵循Django文档，当View由以下内容定义时，分页系统会起作用：def view(request):

我也想知道如何取出下一个和之前的对象。

我需要一个指向“下一期（带有名称和发行号码的上下文）”的链接，然后是“上一期”链接。请注意，只需使用下一个或上一个问题更改模板链接就无法运行。

所以，如果有人能帮助我，那就太好了。

Answer 1

只需将paginate_by = 20添加到您的观看类。

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

    #More stuff here..

Answer 2

就像Evan Porter所评论的那样，您可以使用page_obj上下文变量来访问number, paginatior.num_pages, has_next, has_previous。从Django 1.4.1升级到1.7，KeyError['page']

后，这就是我从object_list to ListView救了我的东西

Django中的Paginate基于类的视图

2 个答案: