Django错误报告 - 如何知道哪个用户触发了错误？

Question

有没有办法我可以自定义Django错误报告,所以当它通过电子邮件发送给我时,它让我知道哪个用户触发了错误？

如果重要的话,我在Django 1.2.

非常感谢提前!

Answer 1

如果您不想使用哨兵,您可以使用这个简单的中间件将用户信息附加到错误邮件:

# source: https://gist.github.com/646372
class ExceptionUserInfoMiddleware(object):
    """
    Adds user details to request context on receiving an exception, so that they show up in the error emails.

    Add to settings.MIDDLEWARE_CLASSES and keep it outermost(i.e. on top if possible). This allows
    it to catch exceptions in other middlewares as well.
    """

    def process_exception(self, request, exception):
        """
        Process the exception.

        :Parameters:
           - `request`: request that caused the exception
           - `exception`: actual exception being raised
        """

        try:
            if request.user.is_authenticated():
                request.META['USERNAME'] = str(request.user.username)
                request.META['USER_EMAIL'] = str(request.user.email)
        except:
            pass

您可以简单地将此类放在Django项目下面的*.py文件中,并添加对它的引用MIDDLEWARE_CLASSES.即如果你把它放在项目根目录(你的settings.py所在的)中的"中间件"文件中,你只需添加即可middleware.ExceptionUserInfoMiddleware.