Django REST Swagger实现指定api参数

为什么要指定swagger的api参数

api的参数有多种类型：

query 参数，如 /users?role=admin

path 参数，如 /users/{id}

header 参数，如 X-MyHeader: Value

body 参数，描述POST，PUT，PATCH请求的body

form 参数，描述 Content-Type of application/x-www-form-urlencoded 和 multipart/form-data 的请求报文body的参数

swagger指定api参数就可以在文档相应的api条目中显示出api的描述、正常输出、异常输出、参数的名称、描述、是否必填、值类型、参数类型对不同的参数类型有不同的显示效果。swagger是可交互的api文档，可以直接填入文档显示的参数的值并发送请求，返回的结果就会在文档中显示。

难点

对 Django REST Swagger < 2 的版本，要指定swagger的api参数非常容易，只要将相关说明以特定格式和yaml格式写在相应api的视图函数的文档字符串（DocStrings）里，swagger就会自动渲染到文档中。比如这样的格式：

def cancel(self, request, id):
 """
 desc: 取消任务，进行中的参与者得到报酬
 ret: msg
 err: 404页面/msg
 input:
 - name: id
 desc: 任务id
 type: string
 required: true
 location: path
 """

但是在2.0版本之后，Django REST Swagger废弃了对yaml文档字符串的支持，不会渲染出任何内容。

一种解决方案

在Django REST framework基于类的api视图中定义filter_class过滤出模型（models）的特定字段，swagger会根据这些字段来渲染。

from django_filters.rest_framework.filterset import FilterSet

class ProductFilter(FilterSet):

 class Meta(object):
 models = models.Product
 fields = (
  'name', 'category', 'id', )

class PurchasedProductsList(generics.ListAPIView):
 """
 Return a list of all the products that the authenticated
 user has ever purchased, with optional filtering.
 """
 model = Product
 serializer_class = ProductSerializer
 filter_class = ProductFilter

 def get_queryset(self):
 user = self.request.user
 return user.purchase_set.all()

这个解决方法只解决了一半问题，只能用在面向模型的api，只能过滤模型的一些字段，而且api参数名与模型字段名不一致时还要额外处理。

启发

查阅Django REST Swagger的文档，Advanced Usage提到，基于类的文档api视图是这样的：

from rest_framework.response import Response
from rest_framework.schemas import SchemaGenerator
from rest_framework.views import APIView
from rest_framework_swagger import renderers

class SwaggerSchemaView(APIView):
 permission_classes = [AllowAny]
 renderer_classes = [
 renderers.OpenAPIRenderer,
 renderers.SwaggerUIRenderer
 ]

 def get(self, request):
 generator = SchemaGenerator()
 schema = generator.get_schema(request=request)

 return Response(schema)

说明文档是根据schema变量来渲染的，所以可以通过重载schema变量，利用yaml包解析出api视图函数的文档字符串中的参数定义赋值给schema变量。

更好的解决方法

创建schema_view.py：

from django.utils.six.moves.urllib import parse as urlparse
from rest_framework.schemas import AutoSchema
import yaml
import coreapi
from rest_framework_swagger.views import get_swagger_view

class CustomSchema(AutoSchema):
 def get_link(self, path, method, base_url):

 view = self.view
 method_name = getattr(view, 'action', method.lower())
 method_docstring = getattr(view, method_name, None).__doc__
 _method_desc = ''

 fields = self.get_path_fields(path, method)

 try:
  a = method_docstring.split('---')
 except:
  fields += self.get_serializer_fields(path, method)
 else:
  yaml_doc = None
  if method_docstring:
  try:
   yaml_doc = yaml.load(a[1])
  except:
   yaml_doc = None

  # Extract schema information from yaml

  if yaml_doc and type(yaml_doc) != str:
  _desc = yaml_doc.get('desc', '')
  _ret = yaml_doc.get('ret', '')
  _err = yaml_doc.get('err', '')
  _method_desc = _desc + '\n<br/>' + 'return: ' + _ret + '<br/>' + 'error: ' + _err
  params = yaml_doc.get('input', [])

  for i in params:
   _name = i.get('name')
   _desc = i.get('desc')
   _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 = a[0]
  fields += self.get_serializer_fields(path, method)

 fields += self.get_pagination_fields(path, method)
 fields += self.get_filter_fields(path, method)

 manual_fields = self.get_manual_fields(path, method)
 fields = self.update_fields(fields, manual_fields)

 if fields and any([field.location in ('form', 'body') for field in fields]):
  encoding = self.get_encoding(path, method)
 else:
  encoding = None

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

 return coreapi.Link(
  url=urlparse.urljoin(base_url, path),
  action=method.lower(),
  encoding=encoding,
  fields=fields,
  description=_method_desc
 )

schema_view = get_swagger_view(title='API')

urls.py中指向schema_view：

from .schema_view import schema_view

urlpatterns = [
 url(r'^v1/api/', include([
 url(r'^doc/', schema_view),
 ])),

然后在需要指定api参数的视图类（如APIView或ModelViewSet）中重载schema：

schema = CustomSchema()

以上这篇Django REST Swagger实现指定api参数就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持。

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发布时间 2020-07-07 12:01:59