Django rest framework 向Django Rest Swagger添加基本Url
我找不到将基本URL添加到Django rest framework 向Django Rest Swagger添加基本Url,django-rest-framework,swagger-ui,django-rest-swagger,Django Rest Framework,Swagger Ui,Django Rest Swagger,我找不到将基本URL添加到djangorest-swagger的方法。我试着加上 SWAGGER_SETTINGS = { "base_path": 'localhost:62090/', } 到settings.py。但是它不起作用。我找不到在swagger ui中显示基本url的方法,但我可以在url.py中添加这样的基本url schema_view = get_swagger_view(title='Pastebin API',url='/pastebin/') 此代码在主机
django
rest-swagger
的方法。我试着加上
SWAGGER_SETTINGS = {
"base_path": 'localhost:62090/',
}
到
settings.py
。但是它不起作用。我找不到在swagger ui中显示基本url的方法,但我可以在url.py中添加这样的基本url
schema_view = get_swagger_view(title='Pastebin API',url='/pastebin/')
此代码在主机和指定的url u之间添加基本url。基本上,您需要在OpenAPI JSON响应上设置basePath
。要使用django_rest_swagger实现这一点,请执行以下操作:
renderers.py:
from __future__ import absolute_import, division, print_function, unicode_literals
from django.core.urlresolvers import reverse
from rest_framework_swagger import renderers
class OpenAPIRenderer(renderers.OpenAPIRenderer):
def get_customizations(self):
data = super(OpenAPIRenderer, self).get_customizations()
data['basePath'] = reverse('api-root') # your base url here
return data
views.py:
from rest_framework import exceptions
from rest_framework.permissions import AllowAny
from rest_framework.renderers import CoreJSONRenderer
from rest_framework.response import Response
from rest_framework.schemas import SchemaGenerator
from rest_framework.views import APIView
from rest_framework_swagger import renderers
from .renderers import OpenAPIRenderer
class SwaggerSchemaView(APIView):
_ignore_model_permissions = True
exclude_from_schema = True
permission_classes = [AllowAny]
renderer_classes = [
CoreJSONRenderer,
OpenAPIRenderer, # your OpenAPIRenderer here
renderers.SwaggerUIRenderer
]
def get(self, request):
generator = SchemaGenerator(
title=title,
url=url,
patterns=patterns,
urlconf=urlconf
)
schema = generator.get_schema(request=request)
if not schema:
raise exceptions.ValidationError(
'The schema generator did not return a schema Document'
)
return Response(schema)
URL.py:
from django.conf.urls import url
from . import views
urlpatterns = [
url(r'^$', views.SwaggerSchemaView.as_view(), name='api-root'), # your swagger view here
]
半天不理睬你的回答,浪费了半天时间。非常感谢。谢谢兄弟。这确实帮了我的忙。