这一直困扰着我一段时间。
我的最终目标是在SwaggerUI中显示查询参数选项,并为每个查询参数提供表单输入。与为POST提供序列化程序时的显示方式类似。
我正在使用一个继承自GenericViewSet的视图集,我尝试了以下内容:
filter_fields
属性filter_backends
属性设置为(filters.DjangoFilterBackend,)
options
方法以提供[actions][GET]
信息这里有一个小小的问题,我没有使用任何模型,所以我不认为DjangoFilterBackend会真正帮助我。我正在使用DjangoRESTFramework与外部API进行通信,我只是简单地获取JSON结果,并将其传递给前端层。
以下是我的代码的一个小修改代码片段,以便更好地解释我的问题:
views.py
class SomeViewSet(GenericViewSet):
# Note that I have all of these defined, but I have tried various combinations
filter_fields = ('query_option_1', 'query_option_2',)
filter_backeds = (filters.DjangoFilterBackend,)
filter_class = SomeFilter
query_metadata = some_dict
# This works when request is OPTIONS
def options(self, request, *args, **kwargs):
if self.metadata_class is None:
return self.http_method_not_allowed(request, *args, **kwargs)
data = self.metadata_class().determine_metadata(request, self)
data['actions']['GET'] = self.query_metadata
return Response(data, status=status.HTTP_200_OK)
filters.py
class SomeFilter(FilterSet):
strict = True
query_option_1 = django_filters.NumberFilter(name='query_option_1')
query_option_2 = django_filters.NumberFilter(name='query_option_2')
class Meta:
fields = ['query_option_1', 'query_option_2']
感谢您的关注,并提前感谢您的回复。
答案 0 :(得分:26)
答案 1 :(得分:20)
好的,对于那些偶然发现这个问题的人,我已经弄明白了。这是相当愚蠢的,我觉得有点愚蠢不知道,但在我的辩护中,它没有明确记录。在DRF文档中或Django REST Swagger存储库中找不到该信息。相反,它是在django-rest-framework-docs下找到的,这就是Django REST Swagger的基础。
要指定您的查询参数作为表单字段显示在SwaggerUI中,您只需这样评论:
def list(self):
"""
param1 -- A first parameter
param2 -- A second parameter
"""
...
swagger会解析你的评论,并会为param1和param2输入一个表单。 --
后面的内容是参数的说明。
答案 2 :(得分:12)
我找到了rest framework swagger docs。 所以我们可以编写参数类型(整数,字符),响应等
三重奏---
是必要的。
@api_view(["POST"])
def foo_view(request):
"""
Your docs
---
# YAML (must be separated by `---`)
type:
name:
required: true
type: string
url:
required: false
type: url
created_at:
required: true
type: string
format: date-time
serializer: .serializers.FooSerializer
omit_serializer: false
parameters_strategy: merge
omit_parameters:
- path
parameters:
- name: name
description: Foobar long description goes here
required: true
type: string
paramType: form
- name: other_foo
paramType: query
- name: other_bar
paramType: query
- name: avatar
type: file
responseMessages:
- code: 401
message: Not authenticated
"""
我们使用mixins类如ModelViewSets
的情况如何。
我们是否需要定义list
函数才能添加文档?
- 没有
我们可以这样做:
class ArticleViewSet(viewsets.ModelViewSet):
"""
Articles.
---
list: #<--- here!!
parameters:
- name: name
description: article title
get_price:
omit_serializer: true
"""
@list_route(methods=['get'])
def get_price(self, request):
pass
答案 3 :(得分:1)
免责声明:我正在使用django_filters
,因此结果可能会有所不同。 django_filters
在DRF ViewSet中使用参数filter_fields
,这与不使用django_filters
可能有所不同。
我从this thread中汲取了灵感,并通过以下方式覆盖了过滤后端中的get_schema_fields()
方法。
settings.py
REST_FRAMEWORK = {
...
'DEFAULT_FILTER_BACKENDS': ('location.of.custom_backend.CustomDjangoFilterBackend')
...
}
custom_backend.py
import coreapi
import coreschema
from django_filters.rest_framework import DjangoFilterBackend
class CustomDjangoFilterBackend(DjangoFilterBackend):
"""
Overrides get_schema_fields() to show filter_fields in Swagger.
"""
def get_schema_fields(self, view):
assert (
coreapi is not None
), "coreapi must be installed to use `get_schema_fields()`"
assert (
coreschema is not None
), "coreschema must be installed to use `get_schema_fields()`"
# append filter fields to existing fields
fields = super().get_schema_fields(view)
if hasattr(view, "filter_fields"):
fields += view.filter_fields
return [
coreapi.Field(
name=field,
location='query',
required=False,
type='string',
) for field in fields
]
答案 4 :(得分:1)
使用openapi(而不是coreapi),我发现的“最简单”方法是通过以下core dev comment:
from rest_framework.schemas.openapi import AutoSchema
class CustomSchema(AutoSchema):
def get_operation(self, path, method):
op = super().get_operation(path, method)
op['parameters'].append({
"name": "foo",
"in": "query",
"required": True,
"description": "What foo does...",
'schema': {'type': 'string'}
})
return op
class MyViewSet(ModelViewSet):
schema = CustomSchema()
def get_queryset(self):
foo = self.request.query_params.get("foo")
if foo:
self.queryset = self.queryset.filter(foo=foo)
return self.queryset
答案 5 :(得分:0)
详细说明@vadimchin的上述答案-这是一个有效的示例。
func collectionView(_ collectionView: UICollectionView, layout collectionViewLayout: UICollectionViewLayout, sizeForItemAt indexPath: IndexPath) -> CGSize {
if collectionView == filterCollectionView {
return CGSize(width: collectionView.frame.size.width / 4, height: collectionView.frame.size.height)
} else {
return CGSize(width: collectionView.frame.size.width, height: collectionView.frame.size.height)
}
}
我在我的应用程序中使用了Viewsets。我必须按照@jarussi的建议实施# requirements.txt
djangorestframework==3.9.3
django-rest-swagger==2.2.0
django==2.2.1
coreapi==2.3.3
。
filter_queryset(self, request, queryset, view)
# models.py
from django.db import models
class Recording(models.Model):
_id = models.AutoField(primary_key=True)
name = models.CharField(max_length=511)
# serializers.py
from models import Recording
from rest_framework import serializers
class RecordingSerializer(serializers.ModelSerializer):
class Meta:
model = Recording
fields = '__all__'
# views.py
from rest_framework import viewsets
from filters import NameFilterBackend
from serializers import RecordingSerializer
class RecordingViewSet(viewsets.ModelViewSet):
serializer_class = RecordingSerializer
queryset = Recording.objects.all()
filter_backends = (NameFilterBackend,)
答案 6 :(得分:0)
请参考this github issue来解决问题。
答案 7 :(得分:0)
如果在过滤器后端使用查询参数,添加get_schema_operation_parameters
方法是最简单的解决方案:
class SimpleFilterBackend(BaseFilterBackend):
def filter_queryset(self, request, queryset, view):
foo = request.query_params.get("foo")
if foo:
queryset = queryset.filter(foo=foo)
return queryset
def get_schema_operation_parameters(self, view):
return [{
"name": "foo",
"in": "query",
"required": True,
"description": "What foo does...",
"schema": {"type": "string"}
}]
class MyViewSet(ModelViewSet):
filter_backends = [SimpleFilterBackend]