Home Backend Development Python Tutorial Directly return the requested method from Django's middleware

Directly return the requested method from Django's middleware

Jun 01, 2018 pm 02:35 PM
django middleware ask

This article mainly introduces the method of returning requests directly from Django's middleware. It has certain reference value. Now I share it with you. Friends in need can refer to it

The example is as follows:

#coding=utf-8
import json
import gevent
from django.http import HttpResponse
from sdsom.web.recorder import get_event_type
from sdsom.web.recorder import get_request_event_info
from sdsom.db.rpcclient import get_db_client
class RecordEventMiddleWare(object) :
 def process_view(self, request, view, args, kwargs) :
 etype = get_event_type(request)
 if not etype :
  return None
 info = get_request_event_info(request, etype)
 info['status'] = "BEGIN"
 try:
  get_db_client().add_event_record(info)
 except :
  return HttpResponse(
   json.dumps({"susscess":0, "message":"记录事件开始到数据库出错"}),
   content_type='application/json'
   )
 return None
Copy after login

As shown in the above code, you need to import the HttpResponse class from the http module of django,

Then when returning, you can jsondump the dictionary content you want to return (if not dumped, the upper layer will handle the error).

Related recommendations:

Detailed explanation of calling static files for Django learning

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