问题
I'm fairly new to Wagtail, and I am in the process of creating a site that will have a Resources (blog) section and I'm not sure how to implement pagination so that there are only 5 posts on each page and the user has to click a number (1, 2, 3, etc.) to go to the next page to see the next 5 posts.
I have this in my template for the pagination section of the resource/blog index page:
<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>
What code do I need to incorporate to make this functional? Thanks in advance.
回答1:
Django provides the module django.core.paginator for this purpose: https://docs.djangoproject.com/en/1.10/topics/pagination/ . Using this within Wagtail is very similar to the examples in the Django documentation - the only real difference is that when you're setting up the Paginator object to be passed to the template, you do that with a get_context method on the page model, instead of a view function. Your model definition will look something like this:
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
Within your template, you can now loop over the items using {% for resource in resources %}, and display the pagination links as follows:
<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>
回答2:
I very much appreciate that you got me here - thanks so much for the assist. I had to make some adjustments to make it work. Here's the model if anyone comes across the same issue:
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
...and here's the 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>
It works like a charm - and adds to the learning curve!
回答3:
In case it's useful to anyone, I wanted this to work as closely as possible to the class-based view ListView, and so I ended up with this:
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)
})
This will give you the same paginator, page_obj, is_paginated and object_list variables in your template that you would get with a normal Django ListView.
(Using python 3, Django 2.1 and Wagtail 2.3.)
来源：https://stackoverflow.com/questions/40365500/pagination-in-wagtail