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[英]Django Rest Framework: How to enable swagger docs for function based views
[英]Django Rest Swagger 2: Is there anyway so far to document parameters for POST requests of FUNCTION based views?
YAML 文檔字符串解析器在 REST Swagger>=2.0 中已棄用
我所做的是覆蓋 SchemaGenerator 類以按照我自己的約定解析視圖的文檔字符串。
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
import yaml
import coreapi
import urlparse
class SchemaGenerator(SchemaGenerator):
def get_link(self, path, method, view):
"""Custom the coreapi using the func.__doc__ .
if __doc__ of the function exist, use the __doc__ building the coreapi. else use the default serializer.
__doc__ in yaml format, eg:
description: the desc of this api.
parameters:
- name: mobile
desc: the mobile number
type: string
required: true
location: form
- name: promotion
desc: the activity id
type: int
required: true
location: form
"""
fields = self.get_path_fields(path, method, view)
yaml_doc = None
if view and view.__doc__:
try:
yaml_doc = yaml.load(view.__doc__)
except:
yaml_doc = None
if yaml_doc and type(yaml_doc) != str:
_method_desc = yaml_doc.get('description', '')
params = yaml_doc.get('parameters', [])
for i in params:
_name = i.get('name')
_desc = i.get('description')
_required = i.get('required', False)
_type = i.get('type', 'string')
_location = i.get('location', 'form')
field = coreapi.Field(
name=_name,
location=_location,
required=_required,
description=_desc,
type=_type
)
fields.append(field)
else:
_method_desc = view.__doc__ if view and view.__doc__ else ''
fields += self.get_serializer_fields(path, method, view)
fields += self.get_pagination_fields(path, method, view)
fields += self.get_filter_fields(path, method, view)
if fields and any([field.location in ('form', 'body') for field in fields]):
encoding = self.get_encoding(path, method, view)
else:
encoding = None
if self.url and path.startswith('/'):
path = path[1:]
return coreapi.Link(
url=urlparse.urljoin(self.url, path),
action=method.lower(),
encoding=encoding,
fields=fields,
description=_method_desc
)
def get_swagger_view(title=None, url=None, patterns=None, urlconf=None):
"""
Returns schema view which renders Swagger/OpenAPI.
"""
class SwaggerSchemaView(APIView):
_ignore_model_permissions = True
exclude_from_schema = True
permission_classes = [AllowAny]
renderer_classes = [
CoreJSONRenderer,
renderers.OpenAPIRenderer,
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)
return SwaggerSchemaView.as_view()
在項目結構中的任何位置創建此模塊。 在project/urls.py
從此模塊導入get_swagger_view
。 然后,從django_rest_swagger
模塊中刪除get_swagger_view
方法。
參考: daimon99 在 REST Swagger 問題中的評論
更新:從django-rest-framework
3.7 版開始,由於上述代碼無法正常工作,存在重大更改,解決方案將是GuillaumeCisco 的評論
您可以使用裝飾器:
from rest_framework.decorators import api_view
然后在你的函數上面使用:
@api_view(['POST'])
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