在django中使用自定义标签实现分页功能

效果演示：

github地址：https://github.com/mncu/django_projects/tree/master/django_projects/pagination_test

本例中总页数为30页，显示页数为12页，当前页的前排页数为4，后排页数为5

将分页分为三种情况：

1   当前页为第1页到第7页的时候，无省略页，且12个位置的内容是不变

2  当前页为第8页到第25页时，位置1与位置2内容不变，当前页一直处于位置7,

3  当前页为第25页到第30页时，位置1与位置2内容不变，位置8到位置12的内容不变，当前页在位置8到位置12之中变换

自定义标签代码：

from django import template

register = template.Library()

@register.assignment_tag
def pagination(current_page,paginator,num_of_displaypages=10,num_of_backpages=4):
 # current_page is a django.core.paginator.Page 's instance
 # paginator is a django.core.paginator.Paginator 's instance
 #
 num_of_frontpages = num_of_displaypages - num_of_backpages -3
 html=''

 # 当总页数小于等于 显示页数 时，则将总页数全部显示
 if paginator.num_pages <= num_of_displaypages :
  for i in range(1,paginator.num_pages+1):
   html+= '
%s 
'%(i,i)
  return html
 # 第一种情况 
 elif current_page.number <= num_of_displaypages-num_of_backpages:
  for i in range(1,num_of_displaypages+1):
   html+= '
%s 
'%(i,i)
  return html
 # 第二种情况
 elif num_of_displaypages-num_of_frontpages <= current_page.number <= paginator.num_pages-num_of_backpages :
  html = '''
   
1
   
...

  '''
  for i in range(current_page.number-num_of_frontpages,current_page.number+num_of_backpages+1):
   html+='
%s'%(i,i)
  return html
 # 第三种情况
 else:
  html = '''
   
1
   
...

  '''
  for i in range(paginator.num_pages-num_of_backpages-num_of_frontpages,paginator.num_pages+1):
   html+='
%s'%(i,i)
  return html

来看html代码

{% load mytags %}



 
 
 
 

 
 

 
 

 
 


{{ current_page.object_list }}

 

看看view函数：

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

# Create your views here.


def index(request):

 obj_list = ['page01','page02','page03','page04','page05','page06','page07','page08','page09','page10',
    'page11','page12','page13','page14','page15','page16','page17','page18','page19','page20',
    'page21','page22','page23','page24','page25','page26','page27','page28','page29','page30',]
 #create a paginator instance
 paginator = Paginator(obj_list,1)

 #Get the page_number of current page
 current_page_num = request.GET.get('page')

 try:
  current_page = paginator.page(current_page_num)
 except PageNotAnInteger:
  # If page is not an integer, deliver first page.
  current_page = paginator.page(1)
 except EmptyPage:
  # If page is out of range (e.g. 9999), deliver last page of results.
  current_page = paginator.page(paginator.num_pages)
 return render(request,'index.html',
     {'current_page': current_page,
     'paginator': paginator

     })

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