Question

我想使用通用django ListView通过此URL显示已登录用户的所有帖子：

#/users/myposts/
url(r'^myposts/', views.MyPostsView.as_view(), name='myposts'),

我的帖子模型如下：

class Post(models.Model):
   person = models.ForeignKey(settings.AUTH_USER_MODEL, on_delete=models.CASCADE)
   domain = models.ForeignKey(Domain, on_delete=models.CASCADE)
   speciality = models.ForeignKey(Speciality, on_delete=models.CASCADE)
   level = models.ManyToManyField(Level)
   date_post = models.DateTimeField()
   title = models.CharField(max_length=200)
   description = models.CharField(max_length=1000)
   is_deleted = models.BooleanField(default=False)
   date_delete = models.DateTimeField(null=True)

我的通用ListView是：

class MyPostsView(generic.ListView):
     template_name = 'posts_list.html'
     userId = None

     def get_queryset(self):
         return Post.objects.get(person=self.request.user.id)

这样做我收到以下错误消息

get() returned more than one Post -- it returned 2

但是当我返回Post.objects.all()时它返回所有帖子并显示它们

Answer 1

使用objects.filter而不是objects.get。

（Django）使用通用ListView过滤显示的对象

1 个答案: