Django REST 框架 Swagger 2.0

Question

很难配置 Swagger UI 以下是解释性很强的文档：https://django-rest-swagger.readthedocs.io/en/latest/

YAML 文档字符串已弃用。有人知道如何从 python 代码中配置 Swagger UI 吗？或者我应该更改什么文件来分组 api 端点，向每个端点添加注释，在 Swagger UI?

中添加查询参数字段

Answer 1

我是这样做到的：

基地urls.py

urlpatterns = [
...
url(r'^api/', include('api.urls', namespace='api')),
url(r'^api-auth/', include('rest_framework.urls', namespace='rest_framework')),
...
]

api.urls.py

urlpatterns = [
url(r'^$', schema_view, name='swagger'),
url(r'^article/(?P<pk>[0-9]+)/$', 
    ArticleDetailApiView.as_view(actions={'get': 'get_article_by_id'}), 
    name='article_detail_id'),
url(r'^article/(?P<name>.+)/(?P<pk>[0-9]+)/$', 
    ArticleDetailApiView.as_view(actions={'get': 'get_article'}), 
    name='article_detail'),
]

api.views.py。在 MyOpenAPIRenderer 中，我更新数据字典以添加描述、查询字段并更新类型或所需的功能。

class MyOpenAPIRenderer(OpenAPIRenderer):
    def add_customizations(self, data):
        super(MyOpenAPIRenderer, self).add_customizations(data)
        data['paths']['/article/{name}/{pk}/']['get'].update(
            {'description': 'Some **description**',
             'parameters': [{'description': 'Add some description',
                             'in': 'path',
                             'name': 'pk',
                             'required': True,
                             'type': 'integer'},
                            {'description': 'Add some description',
                             'in': 'path',
                             'name': 'name',
                             'required': True,
                             'type': 'string'},
                            {'description': 'Add some description',
                             'in': 'query',
                             'name': 'a_query_param',
                             'required': True,
                             'type': 'boolean'},
                            ]
             })
        # data['paths']['/article/{pk}/']['get'].update({...})
        data['basePath'] = '/api'  

@api_view()
@renderer_classes([MyOpenAPIRenderer, SwaggerUIRenderer])
def schema_view(request):
    generator = SchemaGenerator(title='A title', urlconf='api.urls')
    schema = generator.get_schema(request=request)
    return Response(schema)


class ArticleDetailApiView(ViewSet):

    @detail_route(renderer_classes=(StaticHTMLRenderer,))
    def get_article_by_id(self, request, pk):
        pass

    @detail_route(renderer_classes=(StaticHTMLRenderer,))
    def get_article(self, request, name, pk):
        pass

django-rest-swagger (2.0.7) 的更新：仅将 add_customizations 替换为 get_customizations.

views.py

class MyOpenAPIRenderer(OpenAPIRenderer):
    def get_customizations(self):
        data = super(MyOpenAPIRenderer, self).get_customizations()
        data['paths'] = custom_data['paths']
        data['info'] = custom_data['info']
        data['basePath'] = custom_data['basePath']
        return data

您可以阅读 swagger specification 创建自定义数据。

Answer 2

所以，似乎发生的事情是 django-rest-frameowrk added the new SchemeGenerator, but it is half-baked and missing the ability to generate action descriptions from code docs, and have an open issue about it，在 3.5.0 中到期。

与此同时，django-rest-swagger 继续更新他们的代码以使用新的 SchemaGenerator，这使其成为 breaking change。

一系列非常奇怪的事件导致了这一点：希望这会尽快得到解决。目前，建议的答案是唯一的选择。

Answer 3

编辑 - 因为 swagger 版本 2.2.0 和 rest 框架 3.9.2 创建了一个自定义模式，如下所示：

from rest_framework.schemas import AutoSchema


class CustomSchema(AutoSchema):
    def get_link(self, path, method, base_url):
        link = super().get_link(path, method, base_url)
        link._fields += self.get_core_fields()
        return link

    def get_core_fields(self):
        return getattr(self.view, 'coreapi_fields', ())

然后，使用 DEFAULT_SCHEMA_CLASS 设置即可。

REST_FRAMEWORK = {
    'DEFAULT_SCHEMA_CLASS': 'common.schema.CustomSchema',
}

！以下方法已过时。

因为我找不到任何可行的选择 here 我只是创建了自己的 SchemaGenerator，如下所示：

from rest_framework.schemas import SchemaGenerator


class MySchemaGenerator(SchemaGenerator):   
    title = 'REST API Index'

    def get_link(self, path, method, view):
        link = super(MySchemaGenerator, self).get_link(path, method, view)
        link._fields += self.get_core_fields(view)
        return link

    def get_core_fields(self, view):
        return getattr(view, 'coreapi_fields', ())

创建了 swagger 视图：

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


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 = MySchemaGenerator()
        schema = generator.get_schema(request=request)

        return Response(schema)

在 urls.py 中使用此视图：

url(r'^docs/$', SwaggerSchemaView.as_view()),

在 APIView 中添加自定义字段：

class EmailValidator(APIView):
    coreapi_fields = (
        coreapi.Field(
            name='email',
            location='query',
            required=True,
            description='Email Address to be validated',
            type='string'
        ),
    )

    def get(self, request):
        return Response('something')

Answer 4

使用建议的解决方案有点老套，但效果很好，实施建议的解决方案可能会遇到一些问题，但此文档解释了 django rest swagger 2 集成以及逐步面临的问题： Django Rest Swagger 2 comprehensive documentation

很晚了，但它可能对现在正在寻求帮助的人有所帮助。

Django REST 框架 Swagger 2.0

Django REST Framework Swagger 2.0

django

django-rest-framework

swagger-ui

swagger-2.0

openapi