Django 错误报告 - 如何知道哪个用户触发了错误？

发布于 2024-11-19 23:38:39 字数 110 浏览 4 评论 0原文

有没有一种方法可以自定义 Django 错误报告，以便当它向我发送电子邮件时，它可以让我知道哪个用户触发了错误？

如果重要的话，我使用的是 Django 1.2。

非常感谢！

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南城旧梦 2024-11-26 23:38:39

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中：

# 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入 Django 项目下方任意位置的 *.py 文件中，并添加对 MIDDLEWARE_CLASSES。即，如果您将其放入项目根目录（settings.py 所在位置）的“中间件”文件中，则只需添加 middleware.ExceptionUserInfoMiddleware 即可。

If you don't want to use sentry you can use this simple middleware to attache the user-infos to the error mail:

# 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

You can simply put this class in a *.py file anywhere below your Django project and add a reference to MIDDLEWARE_CLASSES. I.e. if you put it in a file "middleware" in the project root (where your settings.py is) , you simply add middleware.ExceptionUserInfoMiddleware.

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