Django Rest Swagger 2：到目前为止是否有记录 POST 基于 FUNCTION 的视图请求的参数？

Question

我正在尝试将 django-rest-swagger==2.1.1 与我现有的使用 djangorestframework==3.5.3 的项目集成。

该项目有一些基于 Class 的视图和一些基于函数的视图。集成swagger后，对于"Class Based views"的POST请求显示输入框（明显有序列化器），但对于"function based views"不显示。这个问题已经被问过好几次了，我尝试了以下解决方案：

Solution1

和其他一些人，但对我的情况不起作用。 'function based views' 是否有任何可能的方法，或者我必须将它们转换为基于 class 的视图？

Answer 1

YAML 文档字符串解析器在 REST Swagger>=2.0 中已弃用
我所做的是覆盖 SchemaGenerator class 以按照我自己的约定解析视图的文档字符串。

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()

在项目结构中的任意位置创建此模块。在 project/urls.py 中从该模块导入 get_swagger_view。然后，从 django_rest_swagger 模块中删除 get_swagger_view 方法。

参考：Comment by daimon99 in REST Swagger Issues

更新：从 django-rest-framework 版本 3.7 开始，由于上述代码无法运行，存在重大更改，解决方案是 Comment by GuillaumeCisco

Answer 2

您可以使用装饰器：

from rest_framework.decorators import api_view

然后在你的函数上方使用：

@api_view(['POST'])

Django Rest Swagger 2：到目前为止是否有记录 POST 基于 FUNCTION 的视图请求的参数？

Django Rest Swagger 2: Is there anyway so far to document parameters for POST requests of FUNCTION based views?

python

django

django-rest-framework

swagger-2.0

django-rest-swagger