Django专门提供了一个paginator模块,实现分页很easy。

下面的例子引用了django官方文档:https://docs.djangoproject.com/en/1.11/topics/pagination/

Paginator实例化需要2个参数,一个是待分页的对象list(需要实现count方法或者__len__方法),另一个是每页数量。 
Paginator对象属性:

count 对象数量 
num_pages 总页数 
page_range 所有页数范围的生成器
>>> from django.core.paginator import Paginator
>>> objects = ['john', 'paul', 'george', 'ringo']
>>> p = Paginator(objects, 2)

>>> p.count
4
>>> p.num_pages
2
>>> type(p.page_range)
<class 'range_iterator'>
>>> p.page_range
range(1, 3)

Paginator对象的重要方法,page(),接受一个页码作为参数,返回一个当前页对象。该对象的几个属性和方法:

object_list 当前页的对象列表 
has_next() 有无下一页 
has_ previous() 有无上一页 
has_other_pages() 有无其他页 
next_page_number() 下一页页码 
previous_page_number() 上一页页码 
start_index() 返回当前页第一个对象的在所有对象中的索引,注意,从1开始 
end_index() 返回当前页最后一个对象在所有对象中的索引,注意,从1开始 
paginator 所关联的Paginator对象
>>> page1 = p.page(1)
>>> page1
<Page 1 of 2>
>>> page1.object_list
['john', 'paul']

>>> page2 = p.page(2)
>>> page2.object_list
['george', 'ringo']
>>> page2.has_next()
False
>>> page2.has_previous()
True
>>> page2.has_other_pages()
True
>>> page2.next_page_number()
Traceback (most recent call last):
...
EmptyPage: That page contains no results
>>> page2.previous_page_number()
1
>>> page2.start_index() # The 1-based index of the first item on this page
3
>>> page2.end_index() # The 1-based index of the last item on this page
4

Pagenator的page方法,若传给一个无页码范围之外都值,则得到一个EmptyPage的异常。

>>> p.page(0)
Traceback (most recent call last):
...
EmptyPage: That page number is less than 1
>>> p.page(3)
Traceback (most recent call last):
...
EmptyPage: That page contains no results

在视图中可以这么用:

# views.py
from django.core.paginator import Paginator, EmptyPage, PageNotAnInteger
from django.shortcuts import render

def listing(request):
    contact_list = Contacts.objects.all()
    paginator = Paginator(contact_list, 25) # 每页显示25个

    page = request.GET.get('page') # 拿到页码参数
    try:
        contacts = paginator.page(page)
    except PageNotAnInteger:
        # 捕获到页码不是整数,返回第一页
        contacts = paginator.page(1)
    except EmptyPage:
        # 页码超出范围,返回最后一页
        contacts = paginator.page(paginator.num_pages)

    return render(request, 'list.html', {'contacts': contacts})

contacts对象是用带有分页信息的QuerySet,用来渲染模板:list.html

{% for contact in contacts %}
    {# Each "contact" is a Contact model object. #}
    {{ contact.full_name|upper }}<br />
    ...
{% endfor %}

<div class="pagination">
    <span class="step-links">
        {% if contacts.has_previous %}
            <a href="?page={{ contacts.previous_page_number }}">previous</a>
        {% endif %}

        <span class="current">
            Page {{ contacts.number }} of {{ contacts.paginator.num_pages }}.
        </span>

        {% if contacts.has_next %}
            <a href="?page={{ contacts.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>

基于模型类的视图,只需给paginate_by赋值(每页个数)就可以了,似乎是更简单:

class BookListView(generic.ListView):
    model = Book
    context_object_name = 'book_list'
    template_name = 'catalog/book_list.html'
    paginate_by = 10

模板是通用的,在模板中,可以调用is_paginated 判断是否有分页信息。在用page_obj对象实现分页。

{% block pagination %}
  {% if is_paginated %}
    <div class="pagination">
      <span class="pagelinks">
        {% if page_obj.has_previous %}
          <a href="{{ request.path }}?page={{ page_obj.previous_page_number }}">previous</a>
        {% endif %}
        <span class="page-current">
          Page{{ page_obj.number }} of {{ page_obj.paginator.num_pages }}.
        </span>
        {% if page_obj.has_next %}
        <a href="{{ request.path }}?page={{ page_obj.next_page_number }}">next</a>
        {% endif %}
      </span>
    </div>
  {% endif %}
{% endblock %}