【发布时间】:2016-11-09 07:48:39
【问题描述】:
我创建了一个 ViewSet 类,其中包含一个重写的 list 方法,如下所示:
from rest_framework.response import Response
from rest_framework import viewsets
class MyViewSet(views.ViewSet):
def list(self, request):
return Response([
{"id": 1},
{"id": 2},
])
如何对这个响应进行分页?
在settings.py 我有以下设置:
REST_FRAMEWORK = {
'DEFAULT_PAGINATION_CLASS': 'LinkHeaderPagination',
'PAGE_SIZE': 10
}
而LinkHeaderPagination 是这样构建的:
from rest_framework import pagination
from rest_framework.response import Response
class LinkHeaderPagination(pagination.PageNumberPagination):
page_size_query_param = 'page_size'
def get_paginated_response(self, data):
next_url = self.get_next_link()
previous_url = self.get_previous_link()
if next_url is not None and previous_url is not None:
link = '<{next_url}>; rel="next", <{previous_url}>; rel="prev"'
elif next_url is not None:
link = '<{next_url}>; rel="next"'
elif previous_url is not None:
link = '<{previous_url}>; rel="prev"'
else:
link = ''
link = link.format(next_url=next_url, previous_url=previous_url)
headers = {'Link': link, 'Count': self.page.paginator.count} if link else {}
return Response(data, headers=headers)
这对ModelViewSets 很有效,因为它们有一个指定的查询集,但是我如何对列表进行分页?
【问题讨论】:
标签: python django pagination django-rest-framework