Question

我有两个类似的类，查询过滤器是县代码DE或NL。

是否可以基于url名称进行对象过滤并仅保留一个类？例如，如果我将浏览器指向

127.0.0.1:8000/germany

django将调用过滤器

feed__country__name =＆＃39; DE＆＃39;

和

127.0.0.1:8000/netherland

将使用

feed__country__name =＆＃39; NL＆＃39;？

我的网址：

 url(r'^netherland/$', NLFeedList.as_view(), name='nl'),
 url(r'^germany/$', DEFeedList.as_view(), name='de'),

视图：

class NLFeedList(PaginationMixin, ListView):

    model = FeedItem
    template_name = 'nl_feed.html'
    context_object_name = 'feed_items'
    paginate_by = 20

    def get_queryset(self):
        items = FeedItem.objects.filter(feed__country__name='NL')

        if self.kwargs.get('category', None):
            return items.category(self.kwargs.get('category'))


        return items

    def get_context_data(self, **kwargs):
        context = super(NLFeedList, self).get_context_data(**kwargs)
        context['categories'] = Category.objects.filter(country__name='NL')
        return context

class DEFeedList(PaginationMixin, ListView):

    model = FeedItem
    template_name = 'de_feed.html'
    context_object_name = 'feed_items'

    def get_queryset(self):
        items = FeedItem.objects.filter(feed__country__name='DE')

        if self.kwargs.get('category', None):
            return items.category(self.kwargs.get('category'))

        return items

    def get_context_data(self, **kwargs):
        context = super(DEFeedList, self).get_context_data(**kwargs)
        context['categories'] = Category.objects.filter(country__name='DE')
        return context

Answer 1

您可以执行以下操作：

<强> urls.py

 url(r'^(?P<country>germany|netherland)/$', FeedList.as_view(), name='feedlist')

和视图：

class FeedList(PaginationMixin, ListView):

    model = FeedItem
    context_object_name = 'feed_items'

    match = {'germany':'DE','netherland':'NL'}

    def get_queryset(self):
        code = self.match[self.kwargs['country']]
        items = FeedItem.objects.filter(feed__country__name=code)
        self.template_name = '%s_feed.html' % code.lower()
        if self.kwargs.get('category', None):
            return items.category(self.kwargs.get('category'))


        return items

    def get_context_data(self, **kwargs):
        context = super(FeedList, self).get_context_data(**kwargs)
        context['categories'] = Category.objects.filter(country__name=self.match[self.kwargs['country']])
        return context

此外，您可能不需要两个模板，只需要一个，在这种情况下，只需删除此行self.template_name = '%s_feed.html' % code.lower()并相应地设置template_name。

Answer 2

将您的网址更改为

url(r'^(?P<country>netherland|germany)/$', NLFeedList.as_view(), name='nl'),

然后使用

在视图中访问此新<country>参数

country = self.kwargs['country']

然后在您的视图中执行必要的if country = '...':代码块。

对象基于url请求进行过滤

2 个答案: