如何包含请求。用户详细信息在Django的生产站点的追溯电子邮件中

Question

我想在发生错误时通过电子邮件发送给站点管理员的上下文详细信息中包含request.user的内容，以及追溯和请求.GET/POST/COOKIES/META

任何帮助表示赞赏。

Answer 1

因为process_exception中间件被传入的请求对象，你可以添加你喜欢的任何信息，以request.META

class ErrorMiddleware(object): 
    """ 
    Alter HttpRequest objects on Error 
    """ 

    def process_exception(self, request, exception): 
     """ 
     Add user details. 
     """ 
     request.META['USER'] = request.user.username 

Answer 2

制作一个具有process_exception方法的中间件。 http://docs.djangoproject.com/en/dev/topics/http/middleware/#process-exception

import sys 
import traceback 
from django.conf import settings 
from django.core.mail import mail_admins 

class ProcessExceptionMiddleware(object): 
    def process_exception(self, request, exception): 
     if not settings.DEBUG: 
      msg = '\n\n'.join([request.user, request.GET, request.POST, \ 
       request.COOKIES, request.META, traceback.format_exc(*sys.exc_info())]) 

      mail_admins("Error!", msg) 

我希望给你一些想法！