Question

我写过

class CustomApiException(APIException):

    #public fields
    detail = None
    status_code = None

    # create constructor
    def __init__(self, status_code, message):
        #override public fields
        CustomApiException.status_code = status_code
        CustomApiException.detail = message
        CustomApiException.message = message

适用于APIView，但对于TemplateView，它会产生错误。编写自定义API异常的正确方法是什么，它将适用于两个视图。

Answer 1

在您的视图中，将模板视图定义为以下

from django.http import HttpResponseForbidden

#rest of the views
class yourview(TemplateView):
    def get(request):
        try:
            #business logic
        except CustomApiException:
            return HttpResponseForbidden()

这会引发Error 403

如果要渲染自己的模板而不是引发HttpResponseForbidden，可以按照以下方式将异常变量传递给视图，并适当地写出视图的响应

class yourview(TemplateView):
    def get(request):
        try:
            #business logic
        except CustomApiException as e:
            status_code = e.status_code
            detail = e.message
            message = e.message
            #Handle the response here.

如何在Django中为TemplateView和APIView编写自定义异常，它将返回请求的自定义错误

1 个答案: