Python:Django3x分页分个明白
作者:互联网
from django.core.paginator import Paginator, PageNotAnInteger, EmptyPage, InvalidPage
def page_results(contacts, page, page_size):
page_size = page_size if type(page_size) is int else int(page_size)
paginator = Paginator(contacts, page_size)
kwargs = {
'per_page': 0,
'has_previous': 0,
'number': 0,
'num_pages': 0,
'previous_page_number': 0,
'next_page_number': 0,
'count': 0,
}
try:
page = page if type(page) is int else int(page)
contacts = paginator.page(page)
kwargs['per_page'] = contacts.paginator.per_page
kwargs['has_previous'] = contacts.has_previous
kwargs['number'] = contacts.number
kwargs['num_pages'] = contacts.paginator.num_pages
kwargs['previous_page_number'] = contacts.previous_page_number
kwargs['next_page_number'] = contacts.next_page_number
kwargs['count'] = contacts.paginator.count
# TODO: 开始捕获异常
except PageNotAnInteger:
contacts = paginator.page(1)
except InvalidPage:
# 如果请求的页数不存在, 重定向页面
contacts = paginator.page(1)
except EmptyPage:
# 如果请求的页数不在合法的页数范围内,返回结果的最后一页。
contacts = paginator.page(paginator.num_pages)
kwargs['per_page'] = contacts.paginator.per_page
kwargs['has_previous'] = contacts.has_previous
kwargs['number'] = contacts.number
kwargs['num_pages'] = contacts.paginator.num_pages
kwargs['previous_page_number'] = contacts.previous_page_number
kwargs['next_page_number'] = contacts.next_page_number
kwargs['count'] = contacts.paginator.count
kwargs['page'] = page
kwargs['page_size'] = page_size
return contacts, kwargs
标签:paginator,分页,contacts,Python,number,Django3x,kwargs,page,previous 来源: https://blog.csdn.net/shunzi2016/article/details/120394653