Django:如何在基于类的视图中获取登录用户的用户名?

Django:如何在基于类的视图中获取登录用户的用户名?,django,django-class-based-views,Django,Django Class Based Views,我已实现DeleteView,并希望它在删除对象后将其重定向到登录用户的配置文件: class DeleteAttendanceFeedItem(DeleteView, LoginRequiredMixin): model = AttendanceFeedItem template_name = "users/delete_attendance_feed_item.html" context_object_name = "attendance

我已实现DeleteView,并希望它在删除对象后将其重定向到登录用户的配置文件:

class DeleteAttendanceFeedItem(DeleteView, LoginRequiredMixin):
    model = AttendanceFeedItem
    template_name = "users/delete_attendance_feed_item.html"
    context_object_name = "attendance_feed_item"
    success_url = reverse_lazy("user_profile", kwargs={"username" : request.user.username})
但是,我无法在基于类的视图中访问请求对象。搜索StackOverflow表明我可以访问self.request.user.username,但这会引发相同的错误


这似乎是一件非常简单的事情,我花了大量时间阅读SO和Django文档,但毫无用处。帮助?

使用
self.request.user.username
获取方法中的用户名

class DeleteAttendanceFeedItem(DeleteView,LoginRequiredMixin):
模型=AttendanceFeedItem
template\u name=“users/delete\u attention\u feed\u item.html”
context\u object\u name=“考勤\u提要\u项目”
def get_success_url(自我):

return reverse\u lazy(“user\u profile”,kwargs={“username”:self.request.user.username})
非常感谢,我知道这会很简单!我确实看到这个方法被引用了,但我试图做一些愚蠢的事情,我定义了这个函数,但随后做了
success\u url=get\u success\u url()
,这使得这不可能。我真的很感激你帮我澄清了这件事!10分钟后我会接受你的回答。
class DeleteAttendanceFeedItem(DeleteView, LoginRequiredMixin):
    model = AttendanceFeedItem
    template_name = "users/delete_attendance_feed_item.html"
    context_object_name = "attendance_feed_item"

    def get_success_url(self):
        return reverse_lazy("user_profile", kwargs={"username": self.request.user.username})