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我正在使用自定义 mixin 将过滤器应用于使用 MultipleObjectMixin 分页的基于类的视图的查询集。当我应用过滤器以使当前页面超出数据集范围时，显然会得到 404。我想要做的是捕获此异常并将其重定向到新的最后一页。

我的 get_queryset 有问题的视图：

def get_queryset(self):
    filters = self.build_location_filter()
    if not filters == None:
        return models.Account.corporateAccounts(locations=self.build_location_filter)
    else:
        return models.Account.corporateAccounts()

有什么建议么？

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1 回答 1

4

尝试：

def get(self, request, *args, **kwargs):
    try:
        return super(MyView, self).get(request, *args, **kwargs)
    except Http404:
        if kwargs['page'] > self.paginator.num_pages:
            return HttpResponseRedirect(reverse('this_view_paged', kwargs={'page': self.paginator.num_pages}))
        else:
            # re-raise Http404, as the reason for the 404 was not that maximum pages was exceeded
            raise Http404(_(u"Empty list and '%(class_name)s.allow_empty' is False.")
                      % {'class_name': self.__class__.__name__})

更新

对不起。我真的以为 Django 添加paginator为实例变量。没有它真的很糟糕，因为它使这比它需要的困难100倍。分页器被隐藏在 中paginate_queryset，这实际上是从那里Http404得到的。这意味着，你甚至不能调用这个方法来获取你的分页器，因为你从中得到的只是一个异常。所以，你必须更深入，不幸的是重复了一些 Django 视图逻辑，我讨厌这样做，但这是我能看到的唯一前进道路。由于我们正在诉诸代码重复，无论如何，我现在要覆盖整个paginate_queryset方法。新代码直接从 Django 源代码复制而来，并在注释部分中注明了修改（我为后代保留了上面的原始代码）：

def paginate_queryset(self, queryset, page_size):
        """
        Paginate the queryset, if needed.
        """
        paginator = self.get_paginator(queryset, page_size, allow_empty_first_page=self.get_allow_empty())
        page = self.kwargs.get('page') or self.request.GET.get('page') or 1
        try:
            page_number = int(page)
        except ValueError:
            if page == 'last':
                page_number = paginator.num_pages
            else:
                raise Http404(_(u"Page is not 'last', nor can it be converted to an int."))
        try:
            page = paginator.page(page_number)
            # Moving this line after the try/except block because DRY
            #return (paginator, page, page.object_list, page.has_other_pages())
        except InvalidPage:
            # This used to raise a 404, but we're replacing this functionality
            #raise Http404(_(u'Invalid page (%(page_number)s)') % {
            #                    'page_number': page_number
            #})
            page = paginator.page(paginator.num_pages)

        return (paginator, page, page.object_list, page.has_other_pages())
于 2012-07-19T21:44:32.043 回答