我正在使用Django REST框架与django-rest-swagger一起为我的API创建可浏览的接口。我可以使用YAML文档串指定请求体序列化程序,但还没有找到为请求查询参数指定序列化程序的方法。我使用的视图非常类似于:
class ListBans(BaseBanView):
def get(self, request):
"""
List all profile bans
---
response_serializer: backend_serializers.BanSerializer
request_serializer: moderator_serializers.ListBansSerializer
"""
serializer = moderator_serializers.ListBansSerializer(data=request.query_params)
if serializer.is_valid(raise_exception=True):
# query profile bans
data = []
return APIResponse(status=status.HTTP_200_OK, data=data)
class ListBansSerializer(serializers.Serializer):
limit = serializers.IntegerField(default=10, help_text='query limit')
offset = serializers.IntegerField(default=0, help_text='query offset')我试图实现的是让django-rest-swagger为来自ListBansSerializer的查询参数创建表单字段,这样我就不必在docstring中手动指定parameters部分了。有办法吗?
发布于 2016-06-14 10:52:58
我已经创建了一个简单的装饰器,可以自动将查询参数附加到docstring。该源可在github上获得。下面是一个用法示例:
class ListBansSerializer(serializers.Serializer):
limit = serializers.IntegerField(default=10, help_text='query limit')
offset = serializers.IntegerField(default=0, help_text='query offset')
class ListBans(BaseBanView):
@add_query_parameters(ListBansSerializer)
def get(self, request):
"""
List all profile bans
---
response_serializer: backend_serializers.BanSerializer
"""
serializer = moderator_serializers.ListBansSerializer(data=request.query_params)
if serializer.is_valid(raise_exception=True):
# query profile bans
data = []
return APIResponse(status=status.HTTP_200_OK, data=data)https://stackoverflow.com/questions/35043432
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