我想使用django-rest-framework创建API。到目前为止,我已设法设置一个API端点并设法获取所有项目。基本响应(没有后面描述的BaseResponse类)看起来像这样:
[
{
"uuid": "1db6a08d-ec63-4beb-8b41-9b042c53ab83",
"created_at": "2018-03-12T19:25:07.073620Z",
"updated_at": "2018-03-12T19:25:37.904350Z",
"deleted_at": null,
"random_name": "random name"
}
]
我想要达到的结果是:
[
"success": true
"message": "Some exception message",
"data" :{
"uuid": "1db6a08d-ec63-4beb-8b41-9b042c53ab83",
"created_at": "2018-03-12T19:25:07.073620Z",
"updated_at": "2018-03-12T19:25:37.904350Z",
"deleted_at": null,
"random_name": "random name"
}
]
我设法通过创建一个BaseReponse
类来实现这一点,在视图中我只返回BaseResponse.to_dict()(我在类中创建的一个方法)。
class BaseResponse(object):
data = None
success = False
message = None
def __init__(self, data, exception):
self.data = data
self.message = str(exception) if exception is not None else None
self.success = exception is None
def to_dict(self):
return {
'success': self.success,
'message': self.message,
'data': self.data,
}
查看:
class RandomModelList(APIView):
def get(self, request, format=None):
exception = None
models = None
try:
models = RandomModel.objects.all()
except Exception as e:
exception = e
serializer = RandomModelSerializer(models, many=True)
base_response = BaseResponse(data=serializer.data, exception=exception)
return Response(base_response.to_dict())
我想提一下,使用当前的代码,一切都按预期工作,但我对代码有一个巨大的双重(我只是觉得我重新发明了轮子)。有人能告诉我这是否是我的问题的最佳解决方案,如果不是我应该更改/使用什么?
答案 0 :(得分:1)
您可以改为创建自定义渲染器。像
这样的东西class CustomRenderer(JSONRenderer):
def render(self, data, accepted_media_type=None, renderer_context=None):
resp = {
'data': data
}
return super(CustomRenderer, self).render(resp, accepted_media_type, renderer_context)
然后创建一个像
这样的中间件class CustomResponseMiddleware:
def __init__(self, get_response):
self.get_response = get_response
def __call__(self, request):
response = self.get_response(request)
response.data.update({'success': is_client_error(response.status_code) or is_server_error(response.status_code)})
return response