Django REST例外

Question

我目前有一些基于Django REST Framework的视图代码.我一直在使用客户异常类,但理想情况下我想使用内置的Django REST异常.

从下面的代码中我觉得这可能不是最好或最干净的方式来利用REST Framework异常.

有没有人有任何好的例子,他们正在捕捉问题并使用REST内置异常干净地返回它们？

class JSONResponse(HttpResponse):
    def __init__(self, data, **kwargs):
       content = JSONRenderer().render(data)
       kwargs['content_type'] = 'application/json'
       super(JSONResponse, self).__init__(content, **kwargs)

def queryInput(request):
    try:
        auth_token = session_id = getAuthHeader(request)
        if not auth_token:
            return JSONResponse({'detail' : "fail", "error" : "No X-Auth-Token Found", "data" : None}, status=500)

        if request.method:
            data = JSONParser().parse(request)
            serializer = queryInputSerializer(data=data)

        if request.method == 'POST':
            if serializer.is_valid():
                input= serializer.data["input"]
                fetchData = MainRunner(input=input,auth_token=auth_token)
                main_data = fetchData.main()

            if main_data:
                return JSONResponse({'detail' : "success", "error" : None, "data" : main_data}, status=201)

        return JSONResponse({'detail' : "Unknown Error","error" : True, "data" : None}, status=500)

    except Exception as e:
           return JSONResponse({'error' : str(e)},status=500)

Answer 1

Django REST框架提供了几个内置的异常,这些异常主要是DRF的子类APIException.

您可以像在Python中一样在视图中引发异常:

from rest_framework.exceptions import APIException

def my_view(request):
    raise APIException("There was a problem!")

您还可以通过继承APIException和设置status_code和创建自己的自定义异常default_detail.一些在建的有:ParseError,AuthenticationFailed,NotAuthenticated,PermissionDenied,NotFound,NotAcceptable,ValidationError,等.

然后,这些将由ResponseREST Framework的异常处理程序转换为a .每个异常都与添加到的状态代码相关联Response.默认情况下,异常处理程序设置为内置处理程序:

REST_FRAMEWORK = {
    'EXCEPTION_HANDLER': 'rest_framework.views.exception_handler'
}

但是,如果要通过在settings.py文件中更改此异常来自行转换异常,则可以将其设置为自己的自定义异常处理程序:

REST_FRAMEWORK = {
    'EXCEPTION_HANDLER': 'my_project.my_app.utils.custom_exception_handler'
}

然后在该位置创建自定义处理程序:

from rest_framework.views import exception_handler

def custom_exception_handler(exc, context):
    # Call REST framework's default exception handler first,
    # to get the standard error response.
    response = exception_handler(exc, context)

    # Now add the HTTP status code to the response.
    if response is not None:
        response.data['status_code'] = response.status_code

    return response

Answer 2

您可以在 DRF 异常中使用构建，只需导入并引发

from rest_framework.exceptions import
...
raise ParseError('I already have a status code!')