鹡鸰中的分页

Posted 2023-03-10

【中文标题】鹡鸰中的分页

【英文标题】：Pagination in Wagtail

【发布时间】：2017-03-14 21:54:18

【问题描述】：

我对 Wagtail 还很陌生，我正在创建一个包含资源（博客）部分的网站，但我不确定如何实现分页，以便每个页面上只有 5 个帖子用户必须单击一个数字（1、2、3 等）才能转到下一页以查看接下来的 5 个帖子。

我在资源/博客索引页面的分页部分的模板中有这个：

<ul class="pagination">
  <li><a href="#"><i class="fa fa-angle-left"></i></a></li>
  <li class="active"><a href="#">1</a></li>
  <li><a href="#">2</a></li>
  <li><a href="#">3</a></li>
  <li><a href="#"><i class="fa fa-angle-right"></i></a></li>
</ul>

我需要合并哪些代码才能使其正常工作？提前致谢。

【参考方案1】：

Django 为此提供了模块django.core.paginator：https://docs.djangoproject.com/en/1.10/topics/pagination/。在 Wagtail 中使用它与 Django 文档中的示例非常相似 - 唯一真正的区别是，当您设置要传递给模板的 Paginator 对象时，您可以使用 get_context 方法在页面模型，而不是视图函数。您的模型定义将如下所示：

from django.core.paginator import Paginator, EmptyPage, PageNotAnInteger

class ResourceIndexPage(Page):
    # ...
    def get_context(self, request):
        context = super(ResourceIndexPage, self).get_context(request)

        # Get the full unpaginated listing of resource pages as a queryset -
        # replace this with your own query as appropriate
        all_resources = ResourcePage.objects.live()

        paginator = Paginator(all_resources, 5) # Show 5 resources per page

        page = request.GET.get('page')
        try:
            resources = paginator.page(page)
        except PageNotAnInteger:
            # If page is not an integer, deliver first page.
            resources = paginator.page(1)
        except EmptyPage:
            # If page is out of range (e.g. 9999), deliver last page of results.
            resources = paginator.page(paginator.num_pages)

        # make the variable 'resources' available on the template
        context['resources'] = resources

        return context

在您的模板中，您现在可以使用% for resource in resources % 循环遍历项目，并显示分页链接，如下所示：

<ul class="pagination">
  % if resources.has_previous %
    <li><a href="?page= resources.previous_page_number "><i class="fa fa-angle-left"></i></a></li>
  % endif %
  % for page_num in resources.paginator.page_range %
    <li % if page_num == resources.number %class="active"% endif %><a href="?page= page_num "> page_num </a></li>
  % endfor %
  % if resources.has_next %
    <li><a href="?page= resources.next_page_number "><i class="fa fa-angle-right"></i></a></li>
  % endif %
</ul>

【讨论】：

只是想补充一点，我在最后一个 <li></li> 周围使用了以下内容，仅在有下一页时才显示下一个箭头 - % if resources.has_next % % endif %

不错。这真的很好用。有什么方法可以更新它，以便我可以使用包含 /page/1/ 的漂亮 url？目前我正在乱搞，但很乱！

@Designer023 查看RoutablePageMixin：docs.wagtail.io/en/v1.13.1/reference/contrib/routablepage.html

【参考方案2】：

非常感谢您让我来到这里 - 非常感谢您的帮助。我必须进行一些调整才能使其正常工作。如果有人遇到同样的问题，这里是模型：

class NewsIndexPage(Page):
intro = RichTextField(blank=True)

def get_context(self, request):
    context = super(NewsIndexPage, self).get_context(request)

    # Get the full unpaginated listing of resource pages as a queryset -
    # replace this with your own query as appropriate
    blogpages = self.get_children().live().order_by('-first_published_at')

    paginator = Paginator(blogpages, 3) # Show 3 resources per page

    page = request.GET.get('page')
    try:
        blogpages = paginator.page(page)
    except PageNotAnInteger:
        # If page is not an integer, deliver first page.
        blogpages = paginator.page(1)
    except EmptyPage:
        # If page is out of range (e.g. 9999), deliver last page of results.
        blogpages = paginator.page(paginator.num_pages)

    # make the variable 'resources' available on the template
    context['blogpages'] = blogpages

    return context

...这是 HTML：

<ul class="pagination">
    % if blogpages.has_previous %
      <li>
        <a href="?page= blogpages.previous_page_number "><i class="fa fa-angle-left"></i></a>
      </li>
    % endif %
    % for page_num in blogpages.paginator.page_range %
      <li % if page_num == blogpages.number % class="active"% endif %>
        <a href="?page= page_num "> page_num </a>
      </li>
    % endfor %
    % if resources.has_next %
      <li>
        <a href="?page= blogpages.next_page_number "><i class="fa fa-angle-right"></i></a>
      </li>
      % endif %
  </ul>

它就像一个魅力 - 并增加了学习曲线！

【参考方案3】：

如果它对任何人都有用，我希望它与class-based view ListView 尽可能接近，所以我最终得到了这个：

from django.core.paginator import Paginator, InvalidPage
from django.http import Http404
from django.utils.translation import gettext as _

from wagtail.core.models import Page

class ArticleListPage(Page):

    # Some Page variables set here. #

    # Pagination variables:
    paginator_class = Paginator
    paginate_by = 10
    page_kwarg = 'page'
    paginate_orphans = 0
    allow_empty = False

    def get_context(self, request):
        context = super().get_context(request)

        queryset = Page.objects.live()

        paginator, page, queryset, is_paginated = self.paginate_queryset(
                                        queryset, self.paginate_by, request)
        context.update(
            'paginator': paginator,
            'page_obj': page,
            'is_paginated': is_paginated,
            'object_list': queryset,
        )

        return context

    def paginate_queryset(self, queryset, page_size, request):
        """
        Adapted from the ListView class-based view.
        Added the request argument.
        """
        paginator = self.paginator_class(
                                    queryset,
                                    self.paginate_by,
                                    orphans=self.paginate_orphans,
                                    allow_empty_first_page=self.allow_empty)
        page_kwarg = self.page_kwarg
        page = request.GET.get(page_kwarg) or 1

        try:
            page_number = int(page)
        except ValueError:
            if page == 'last':
                page_number = paginator.num_pages
            else:
                raise Http404(_("Page is not 'last', nor can it be converted to an int."))
        try:
            page = paginator.page(page_number)
            return (paginator, page, page.object_list, page.has_other_pages())
        except InvalidPage as e:
            raise Http404(_('Invalid page (%(page_number)s): %(message)s') % 
                'page_number': page_number,
                'message': str(e)
            )

这将在您的模板中为您提供与普通 Django ListView 相同的 paginator、page_obj、is_paginated 和 object_list 变量。

（使用 python 3、Django 2.1 和 Wagtail 2.3。）