python - 如何在基于 Django 类的 View 中访问当前用户

Question

我无法在基于 Django 类的 View 中访问当前登录的用户:

模型.py:

class Userproject(models.Model):
    class Meta:
        verbose_name = u'pp'
        verbose_name_plural = u'pps'

    user = models.ForeignKey(settings.AUTH_USER_MODEL, 
    related_name="project", verbose_name=_("Владелец проекта"))
    #user = models.ForeignKey(User, unique=True)

    name = models.TextField(u'Название проекта', unique=True)
    date_created = models.DateTimeField(u'Дата создания', 
        default=datetime.now(), db_index=True)
    date_until = models.DateTimeField(u'Оплачен по', default=datetime.now(), db_index=True)

views.py:

@login_required
class UserprojectList(ListView):
    context_object_name = 'userproject_list'
    queryset = Userproject.objects.filter(user=self.request.user)
    template_name = 'userproject_list.html'  

当我导航到 url 时，我看到错误:未定义名称“self”

如果我将 self.request.user 更改为 request.user错误是:name 'request' is not defined

请注意，在没有用户过滤的情况下， View 可以正常工作并显示数据 Django 1.8.5

Answer 1

你可以覆盖get_queryset:

@login_required
class UserprojectList(ListView):
    context_object_name = 'userproject_list'
    template_name = 'userproject_list.html'
    def get_queryset(self):
        return Userproject.objects.filter(user=self.request.user)

另外你不能在类上使用装饰器，所以你必须这样写:

from django.utils.decorators import method_decorator

class UserprojectList(ListView):
    context_object_name = 'userproject_list'
    template_name = 'userproject_list.html'

    @method_decorator(login_required)
    def dispatch(self, *args, **kwargs):
        return super(UserprojectList, self).dispatch(*args, **kwargs)

    def get_queryset(self):
        return Userproject.objects.filter(user=self.request.user)