Django使用django-nonrel在GAE上发出信号

问题描述

我在我的GAE项目上使用django-nonrel.我的要求是，在我的应用程序中，一次只能有一个用户使用给定的用户名登录.我尝试实施以下建议的方法:在Django应用中，每个用户只能进行一次并发登录和如何我可以检测到来自不同位置的Django Web应用程序的多次登录吗?但是问题是这两种方法都可以在开发服务器上使用，但不适用于Google App引擎.因此，我选择了django-signals作为替代方法.我创建了一个post_login信号，该信号会将每个登录用户的用户名存储在数据库的Visitor表中.每次注销时，其他信号post_logout将从该表中删除该用户.部分代码如下:

I am using django-nonrel for my project on GAE. My requirement is that in my application at a time only one user should login with the given username. I tried to implement the following suggested approaches:Allow only one concurrent login per user in django app and How can I detect multiple logins into a Django web application from different locations?But the problem is that both of the approaches working on the development server but didn't work on google app engine. So I switched to django-signals as my alternate approach. I created one post_login signal which will store the username for every login user in a table Visitor in database. On every logout,other signal post_logout will remove the user from this table.The part of codes are as:

#signals.py
post_login = django.dispatch.Signal(providing_args=['request', 'user'])
post_logout = django.dispatch.Signal(providing_args=['request', 'user'])
#models.py
def login_handler(sender,user, **kwargs):
    try:
        result=Visitor.objects.get(user=user)
        print "You already have login with your name"
    except:
        visitor=Visitor()
        visitor.user=user
        visitor.save()
post_login.connect(login_handler)


def logout_handler(sender,user, **kwargs):
    try:
        result=Visitor.objects.get(user=user)
        result.delete()
    except:
        return False
post_logout.connect(logout_handler)

#django.contrib.auth.__init.py__
def login(request):
 :
 user_logged_in.send(sender=user.__class__, request=request, user=user)
  post_login.send(sender=None,request=request, user=user)

def logout(request):
:
user_logged_out.send(sender=user.__class__, request=request, user=user)
post_logout.send(sender=None,request=request, user=user)

请注意，在Google App Engine上运行我的应用程序时出现以下错误.错误:服务器错误服务器遇到错误，无法完成您的请求.

Please note that I am getting the following error while running my application on google app engine.Error: Server ErrorThe server encountered an error and could not complete your request.

我也无法登录该应用程序的管理"部分.请帮助我找到实施此要求的正确方法，或者让我知道我做错了什么.感谢您耐心阅读这个巨大的问题描述:-)

Also I am not able to login into Admin part of the application. Please help me to find right approach to implement this requirement or let me know where I am doing wrong.Thanks for your patience for reading this huge problem description :-)

推荐答案

1.

您不应该像正在做的那样编辑django框架.请勿触摸django.contrib.auth

1.

You should not be editing the django framework like you are doing. Don't touch the files inside django.contrib.auth

如果您希望在某人登录后发送信号，请在您登录该人的视图中发送信号

If you wish to send a signal after someone is logged in, then send the signal in your view where you log the person in

不确定您的实际错误是因为没有显示它(如果这是一个开发环境，则设置DEBUG = True以获得更好的堆栈跟踪)，但是通过查看您的代码，您并没有在信号中正确地获取参数.处理程序.它应该看起来像这样:

Not sure what your actual error is because you are not displaying it (if this is a dev environment set DEBUG = True to get a better stack trace) But by lookingat you code, you are not grabbing the arguments correctly in the signal handler. It should look more like this:

def login_handler(sender, **kwargs):
    try:
        user = kwargs['user']
        request = kwargs['request']
        result=Visitor.objects.get(user=user)
        print "You already have login with your name"
    except:
        visitor=Visitor()
        visitor.user=user
        visitor.save()
post_login.connect(login_handler)

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