Django：当用户登录时发出信号？ [英] Django: signal when user logs in?

问题描述

在我的Django应用程序中，当用户登录时，我需要开始运行一些定期的后台作业，并在用户注销时停止运行，所以我正在寻找一种优雅的方式来

In my Django app, I need to start running a few periodic background jobs when a user logs in and stop running them when the user logs out, so I am looking for an elegant way to

1. 获取用户登录/注销的通知


2. 查询用户登录状态

从我的角度来看，理想的解决方案将是每个<$ c $发送的信号

From my perspective, the ideal solution would be

< c> django.contrib.auth.views.login 和 ... views.logout

1. a方法 django.contrib.auth.models.User.is_logged_in（），类似于 ... User.is_active（）或 ... User.is_authenticated（）

1. a signal sent by each django.contrib.auth.views.login and ... views.logout
2. a method django.contrib.auth.models.User.is_logged_in(), analogous to ... User.is_active() or ... User.is_authenticated()

Django 1.1.1没有这个，我不愿修补来源，并添加它（不知道该怎么做，反正）。

Django 1.1.1 does not have that and I am reluctant to patch the source and add it (not sure how to do that, anyway).

作为一个临时解决方案，我添加了一个默认情况下清除的UserProfile模型的 is_logged_in 布尔字段设置为第一次用户hi登陆页面（由 LOGIN_REDIRECT_URL ='/'定义），并在后续请求中进行查询。我把它添加到UserProfile中，所以我不需要派生和自定义内置的用户模型。

As a temporary solution, I have added an is_logged_in boolean field to the UserProfile model which is cleared by default, is set the first time the user hits the landing page (defined by LOGIN_REDIRECT_URL = '/') and is queried in subsequent requests. I added it to UserProfile, so I don't have to derive from and customize the builtin User model for that purpose only.

我不喜欢这个解决方案。如果用户明确点击注销按钮，我可以清除该标志，但大多数时候，用户只需离开页面或关闭浏览器;在这些情况下清理国旗似乎并不直截了当。另外（虽然数据模型的清晰度，但是）， is_logged_in 不属于UserProfile，而在User模型中。

I don't like this solution. If the user explicitely clicks the logout button, I can clear the flag, but most of the time, users just leave the page or close the browser; clearing the flag in these cases does not seem straight forward to me. Besides (that's rather data model clarity nitpicking, though), is_logged_in does not belong in the UserProfile, but in the User model.

任何人都可以想到替代方法？

Can anyone think of alternate approaches ?

推荐答案

你可以使用这样的信号（我把我的模型。 py）

You can use a signal like this (I put mine in models.py)

  from django.contrib.auth.signals import user_logged_in


  def do_stuff(sender, user, request, **kwargs):
      whatever...

  user_logged_in.connect(do_stuff)

请参阅django docs： https://docs.djangoproject.com/en/dev/ref/contrib/auth/#module-django.contrib.auth.signals 和这里 http://docs.djangoproject.com/en/dev/topics/signals/

See django docs: https://docs.djangoproject.com/en/dev/ref/contrib/auth/#module-django.contrib.auth.signals and here http://docs.djangoproject.com/en/dev/topics/signals/

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