如何在 Django 中更改 JsonResponse 的状态

我的 API 在出错时返回一个 JSON 对象，但是状态代码是 HTTP 200:

response = JsonResponse({'status': 'false', 'message': message})
return response

如何更改响应代码以指示错误？

65100 次浏览
小开

Return an actual status

JsonResponse(status=404, data={'status':'false','message':message})
小开
最佳答案

JsonResponse normally returns HTTP 200, which is the status code for 'OK'. In order to indicate an error, you can add an HTTP status code to JsonResponse as it is a subclass of HttpResponse:

response = JsonResponse({'status':'false','message':message}, status=500)
小开

To change status code in JsonResponse you can do this :

response = JsonResponse({'status':'false','message':message})
response.status_code = 500
return response
小开

Python built-in http library has new class called HTTPStatus which is come from Python 3.5 onward. You can use it when define a status.

from http import HTTPStatus
response = JsonResponse({'status':'false','message':message}, status=HTTPStatus.INTERNAL_SERVER_ERROR)

The value of HTTPStatus.INTERNAL_SERVER_ERROR.value is 500. When someone read your code it's better define someting like HTTPStatus.<STATUS_NAME> other than define an integer value like 500. You can view all the IANA-registered status codes from python library here.

小开

This answer from Sayse works but it's undocumented. If you look at the source you find that it passes the remaining **kwargs to the superclass constructor, HttpStatus. However in the docstring they don't mention that. I don't know if it's the convention to assume that keyword args will be passed to the superclass constructor.

You can also use it like this:

JsonResponse({"error": "not found"}, status=404)

I made a wrapper:

from django.http.response import JsonResponse


class JsonResponseWithStatus(JsonResponse):
"""
A JSON response object with the status as the second argument.


JsonResponse passes remaining keyword arguments to the constructor of the superclass,
HttpResponse. It isn't in the docstring but can be seen by looking at the Django
source.
"""
def __init__(self, data, status=None, encoder=DjangoJSONEncoder,
safe=True, json_dumps_params=None, **kwargs):
super().__init__(data, encoder, safe, json_dumps_params, status=status, **kwargs)

提交答案