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我第一次使用基于类的视图。我很难理解如何使用基于类的视图来实现django-endless-pagination twitter 样式分页。

我可以举一个例子来说明如何去做吗？

这是我的看法：

class EntryDetail(DetailView):
    """
    Render a "detail" view of an object.
    By default this is a model instance looked up from `self.queryset`, but the
    view will support display of *any* object by overriding `self.get_object()`.
    """
    context_object_name = 'entry'
    template_name = "blog/entry.html"
    slug_field = 'slug'
    slug_url_kwarg = 'slug'

    def get_object(self, query_set=None):
        """
        Returns the object the view is displaying.

        By default this requires `self.queryset` and a `pk` or `slug` argument
        in the URLconf, but subclasses can override this to return any object.
        """
        slug = self.kwargs.get(self.slug_url_kwarg, None)
        return get_object_or_404(Entry, slug=slug)
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1 回答 1

4

由于这是一个广泛的问题，我现在想结合几种分页解决方案。

1.使用通用ListView：

from django.views.generic import ListView

class EntryList(ListView):
    model = Entry
    template_name = 'blog/entry_list.html'
    context_object_name = 'entry_list'
    paginate_by = 10

仅使用会更快urls.py：

url(r'^entries/$', ListView.as_view(model=Entry, paginate_by=10))

所以基本上你在这个解决方案中不需要 django-endless-pagination 。您可以在此处查看模板示例：How do I use pagination with Django class based generic ListViews?

2.使用 django-endless-pagination 的AjaxListView：

from endless_pagination.views import AjaxListView    
class EntryList(AjaxListView):
    model = Entry
    context_object_name = 'entry_list'
    page_template = 'entry.html'

或者更快（再次）urls.py仅使用：

from endless_pagination.views import AjaxListView

url(r'^entries/$', AjaxListView.as_view(model=Entry))

参考：http ://django-endless-pagination.readthedocs.org/en/latest/generic_views.html

如果有人知道不同的解决方案，请发表评论。

于 2013-08-02T12:14:58.053 回答