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 %}
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