Django Rest Swagger 2:到目前为止，是否有可以记录基于函数视图的后请求的参数

Question

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。

该项目具有一些基于类的视图和一些基于函数的视图。集成了Swagger之后，它将显示"基于类的视图"（显然具有序列化器）的发布框，但没有显示"基于函数的视图"。这个问题已经多次问，我尝试了遵循解决方案：

解决方案1解决方案2

也很少有其他人，但对我的案件不起作用。是否有任何可能的方法来实现"基于函数的视图"，或者我必须将它们转换为基于类的视图？

Answer 1

yaml docstring解析器在休息swagger> = 2.0
中被弃用我所做的是覆盖示意图类，以通过我自己的惯例来解析视图的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()

在项目结构中的任何位置创建此模块。从project/urls.py中的此模块导入get_swagger_view。然后，然后从django_rest_swagger模块中删除get_swagger_view方法。

参考：daimon99在REST Skagger问题中发表评论

更新：从django-rest-framework版本3.7中，由于上述代码无法正常工作，因此有一些破坏的更改，该解决方案将由Guillaumecisco评论

Answer 2

您可以使用装饰器：

from rest_framework.decorators import api_view

，然后在您的功能上方使用：

@api_view(['POST'])