Django:Paginator错误：类型为'；用户'；没有len（）

我正在学习使用分页器，因为我的列表越来越长了！我在实现它时遇到了问题，因为我遇到了一个错误。我注意到，如果我删除

paginate_by=10

，错误就会消失，但我非常确定这是分页所必需的。为什么get_queryset与paginator冲突

class NotificationsListView(ListView):
    template_name = "notices/list.html"
    paginate_by = 10

    def get_context_data(self, *args, **kwargs):
        context = super(NotificationsListView, self).get_context_data(*args, **kwargs)
        qs = self.get_queryset().notifications.all()
        paginator = Paginator(qs , self.paginate_by)
        page = self.request.GET.get('page')
        try:
            notification_pages = paginator.page(page)
        except PageNotAnInteger:
            notification_pages = paginator.page(1)
        except EmptyPage:
            notification_pages = paginator.page(paginator.num_pages)
        context['notifications'] = notification_pages
            return context    
        def get_queryset(self, *args, **kwargs):
        request = self.request
        return User.objects.get(pk=self.request.user.pk)

回溯：

File "myapp\lib\site-packages\django\core\paginator.py" in count
  85.             return self.object_list.count()

During handling of the above exception ('User' object has no attribute 'count'), another exception occurred:

File "myapp\lib\site-packages\django\core\handlers\exception.py" in inner
  35.             response = get_response(request)

File "myapp\lib\site-packages\django\core\handlers\base.py" in _get_response
  128.                 response = self.process_exception_by_middleware(e, request)

File "myapp\lib\site-packages\django\core\handlers\base.py" in _get_response
  126.                 response = wrapped_callback(request, *callback_args, **callback_kwargs)

File "myapp\lib\site-packages\django\views\generic\base.py" in view
  69.             return self.dispatch(request, *args, **kwargs)

File "myapp\lib\site-packages\django\views\generic\base.py" in dispatch
  89.         return handler(request, *args, **kwargs)

File "myapp\lib\site-packages\django\views\generic\list.py" in get
  157.         context = self.get_context_data()

File "myapp\src\notices\views.py" in get_context_data
  18.         context = super(NotificationsListView, self).get_context_data(*args, **kwargs)

File "myapp\lib\site-packages\django\views\generic\list.py" in get_context_data
  119.             paginator, page, queryset, is_paginated = self.paginate_queryset(queryset, page_size)

File "myapp\lib\site-packages\django\views\generic\list.py" in paginate_queryset
  69.             page = paginator.page(page_number)

File "myapp\lib\site-packages\django\core\paginator.py" in page
  65.         number = self.validate_number(number)

File "myapp\lib\site-packages\django\core\paginator.py" in validate_number
  43.         if number > self.num_pages:

File "myapp\lib\site-packages\django\utils\functional.py" in __get__
  36.         res = instance.__dict__[self.name] = self.func(instance)

File "myapp\lib\site-packages\django\core\paginator.py" in num_pages
  95.         if self.count == 0 and not self.allow_empty_first_page:

File "myapp\lib\site-packages\django\utils\functional.py" in __get__
  36.         res = instance.__dict__[self.name] = self.func(instance)

File "myapp\lib\site-packages\django\core\paginator.py" in count
  90.             return len(self.object_list)

Exception Type: TypeError at /notices/
Exception Value: object of type 'User' has no len()

---

get\u queryset

应返回

queryset

而不是

对象

还有另一个方法叫做
get\u object
，它返回一个对象。
get\u queryset
应该返回一个
queryset
而不是
对象

还有另一个方法叫做
get\u object
，它返回一个对象。
您想显示特定用户的通知列表，对吗？然后，您应该从
get\u queryset
方法返回通知列表

我认为您应该这样做，在我看来，您不应该自己处理分页，django/django restframework应该在内部处理它。

class NotificationsListView(ListView):
    template_name = "notices/list.html"
    paginate_by = 10

    def get_queryset(self, *args, **kwargs):
        return self.request.user.notifications.all()

使用Django Rest框架，您只需要在设置文件中添加默认分页器，类似于这样

REST_FRAMEWORK = {
    'DEFAULT_PAGINATION_CLASS': 'rest_framework.pagination.PageNumberPagination',
    'PAGE_SIZE': 50,
}

或者，如果要使用现有代码，可以尝试以下方法：

def listing(request):
    notifications = request.user.notifications.all()
    paginator = Paginator(notifications, 20) # Show 20 notifications per page

    page = request.GET.get('page')
    try:
        typesets = paginator.page(page)
    except PageNotAnInteger:
        # If page is not an integer, deliver first page.
        typesets = paginator.page(1)
    except EmptyPage:
        # If page is out of range (e.g. 9999), deliver last page of results.
        typesets = paginator.page(paginator.num_pages)

    return render(request, 'list.html', {'typesets': typesets})

您想显示特定用户的通知列表，对吗？然后，您应该从
get\u queryset
方法返回通知列表

我认为您应该这样做，在我看来，您不应该自己处理分页，django/django restframework应该在内部处理它。

class NotificationsListView(ListView):
    template_name = "notices/list.html"
    paginate_by = 10

    def get_queryset(self, *args, **kwargs):
        return self.request.user.notifications.all()

使用Django Rest框架，您只需要在设置文件中添加默认分页器，类似于这样

REST_FRAMEWORK = {
    'DEFAULT_PAGINATION_CLASS': 'rest_framework.pagination.PageNumberPagination',
    'PAGE_SIZE': 50,
}

或者，如果要使用现有代码，可以尝试以下方法：

def listing(request):
    notifications = request.user.notifications.all()
    paginator = Paginator(notifications, 20) # Show 20 notifications per page

    page = request.GET.get('page')
    try:
        typesets = paginator.page(page)
    except PageNotAnInteger:
        # If page is not an integer, deliver first page.
        typesets = paginator.page(1)
    except EmptyPage:
        # If page is out of range (e.g. 9999), deliver last page of results.
        typesets = paginator.page(paginator.num_pages)

    return render(request, 'list.html', {'typesets': typesets})

你能解决你问题中的缩进问题吗？由此可知，
qs
的类型是什么？这是一个
QuerySet
？你能解决你问题中的缩进问题吗？那么，这里的
qs
是什么类型的？那是一个
查询集
？