如何在Django REST Swagger中生成响应消息列表?

时间:2016-10-21 11:23:09

标签: django rest django-rest-framework django-swagger

我昨天将Django REST Framework升级到3.5.0因为我需要很好的模式生成。

我正在使用 Django REST Swagger 来记录我的API,但不知道如何列出API端点提供的所有可能的响应消息。

似乎自动生成与我的终端正在执行的操作相对应的成功消息。

所以POST动作生成201响应代码,没有任何描述。

enter image description here

如何添加我的端点提供的所有响应消息并为其提供一些说明?

我正在使用

djangorestframework==3.5.0

django-rest-swagger==2.0.7

1 个答案:

答案 0 :(得分:20)

啊,终于明白了。

但是!这是黑客攻击 - 可能drf + drf swagger不支持;基本上问题与drf和drf swagger代码没有关系,而是openapi编解码器,看看你自己:

def _get_responses(link):
    """
    Returns minimally acceptable responses object based
    on action / method type.
    """
    template = {'description': ''}
    if link.action.lower() == 'post':
        return {'201': template}
    if link.action.lower() == 'delete':
        return {'204': template}
    return {'200': template}

上述代码位于:openapi_codec/encode.py - github 这与drf或drf swagger没有任何关联 - 只为每个链接(例如:GET / api / v1 / test /)创建一个空描述的模板。

当然,有可能克服这个问题。但正如我所说 - 这是黑客攻击:)我将与你分享一个例子:

docs_swagger.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.views import APIView
from rest_framework_swagger import renderers

from docs_swagger.schema_generator import CustomSchemaGenerator


def get_swagger_view(title=None, url=None):
    """
    Returns schema view which renders Swagger/OpenAPI.

    (Replace with DRF get_schema_view shortcut in 3.5)
    """
    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 = CustomSchemaGenerator(title=title, url=url)  # this is altered line
            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()

我在CustomSchemaGenerator中的操作如下:

docs_swagger.schema_g​​enerator.py

import urlparse
import coreapi
from rest_framework.schemas import SchemaGenerator

from openapi_codec import encode


def _custom_get_responses(link):
    detail = False
    if '{id}' in link.url:
        detail = True
    return link._responses_docs.get(
        '{}_{}'.format(link.action, 'list' if not detail else 'detail'),
        link._responses_docs
    )


# Very nasty; Monkey patching;
encode._get_responses = _custom_get_responses


class CustomSchemaGenerator(SchemaGenerator):

    def get_link(self, path, method, view):
        """
        Return a `coreapi.Link` instance for the given endpoint.
        """
        fields = self.get_path_fields(path, method, view)
        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

        description = self.get_description(path, method, view)

        if self.url and path.startswith('/'):
            path = path[1:]

        # CUSTOM
        data_link = coreapi.Link(
            url=urlparse.urljoin(self.url, path),
            action=method.lower(),
            encoding=encoding,
            fields=fields,
            description=description
        )

        data_link._responses_docs = self.get_response_docs(path, method, view)

        return data_link

    def get_response_docs(self, path, method, view):
        return view.responses_docs if hasattr(view, 'responses_docs') else {'200': {
            'description': 'No response docs definition found.'}
        }

最后:

my_view.py

class TestViewSet(viewsets.ModelViewSet):
    queryset = Test.objects.all()
    serializer_class = TestSerializer

    responses_docs = {
        'get_list': {
            '200': {
                'description': 'Return the list of the Test objects.',
                'schema': {
                    'type': 'array',
                    'items': {
                        'type': 'object',
                        'properties': {
                            'id': {
                                'type': 'integer'
                            }
                        }
                    }
                }
            },
            '404': {
                'description': 'Not found',
                'schema': {
                    'type': 'object',
                    'properties': {
                        'message': {
                            'type': 'string'
                        }
                    }
                },
                'example': {
                    'message': 'Not found.'
                }
            }
        },
        'get_detail': {
            '200': {
                'description': 'Return single Test object.',
                'schema': {
                    'type': 'object',
                    'properties': {
                        'id': {
                            'type': 'integer'
                        }
                    }
                }
            },
            '404': {
                'description': 'Not found.',
                'schema': {
                    'type': 'object',
                    'properties': {
                        'message': {
                            'type': 'string'
                        }
                    }
                },
                'example': {
                    'message': 'Not found.'
                }
            }
        }
    }

我认为这更像是有趣而不是真正的解决方案。在当前状态下可能无法实现真正​​的解决方案。也许你应该问drf swagger的创造者 - 他们是否有计划支持回应?

无论如何,招摇的UI: enter image description here

快乐编码:)