Python Django Rest Swagger 2:到目前为止,是否还有其他方法来记录基于函数视图的POST请求的参数?
我正在尝试将Python Django Rest Swagger 2:到目前为止,是否还有其他方法来记录基于函数视图的POST请求的参数?,python,django,django-rest-framework,swagger-2.0,django-rest-swagger,Python,Django,Django Rest Framework,Swagger 2.0,Django Rest Swagger,我正在尝试将django rest-swagger==2.1.1与使用djangorestframework==3.5.3的现有项目集成 该项目有一些基于类的视图和一些基于函数的视图。集成swagger后,它显示“基于类的视图”(显然有序列化程序)的POST请求的输入框,但不显示“基于函数的视图”。这个问题被问了好几次,我尝试了以下解决方案: 其他人也很少,但对我的案子不起作用。对于“基于函数的视图”,是否有任何可能的方法可以做到这一点,或者我必须将它们转换为基于类的视图?在REST Swa
django rest-swagger==2.1.1
与使用djangorestframework==3.5.3
的现有项目集成
该项目有一些基于类的视图和一些基于函数的视图。集成swagger后,它显示“基于类的视图”(显然有序列化程序)的POST请求的输入框,但不显示“基于函数的视图”。这个问题被问了好几次,我尝试了以下解决方案:
其他人也很少,但对我的案子不起作用。对于“基于函数的视图”,是否有任何可能的方法可以做到这一点,或者我必须将它们转换为基于类的视图?在REST Swagger>=2.0中,YAML docstring解析器被弃用
我所做的是重写SchemaGenerator类,按照我自己的约定解析视图的docstring
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()
在项目结构中的任意位置创建此模块。从project/url.py
中的此模块导入get\u swagger\u view
。然后从django\u rest\u swagger
模块中删除get\u swagger\u view
方法
参考:
更新:从
django rest framework
3.7版开始,有一些突破性的更改,由于这些更改,上述代码无法工作,解决方案将是您可以使用decorator:
from rest_framework.decorators import api_view
然后在你的函数上面使用:
@api_view(['POST'])
全面的答案,但是你的答案显示的示例YAML有一点问题,'int'不起作用,我用字符串替换了它,效果很好,有什么评论吗?哦。我从未真正测试过
类型。谢谢你这么说。我将检查可能出现的问题。无论如何,这对API文档没有帮助。