Django:将空结果页面重定向到根页面的最佳方法
我有一个在Django列表视图中分页和显示对象的应用程序。我正在列表视图中使用Django:将空结果页面重定向到根页面的最佳方法,django,listview,http-status-code-404,Django,Listview,Http Status Code 404,我有一个在Django列表视图中分页和显示对象的应用程序。我正在列表视图中使用paginate_by=20 当我有20多个对象和Google索引时,第二个结果页面/results/?page=2。当结果低于20个对象时,第二页转到404 当第2页、第3页等没有返回任何结果时,将第2页重定向到/results/的最佳方式是什么 我想写一个自定义的404视图来去掉URL参数,但是我想最好在列表视图中捕捉到这个 这是我的列表视图: class ListingListView(ListView):
paginate_by=20
当我有20多个对象和Google索引时,第二个结果页面/results/?page=2
。当结果低于20个对象时,第二页转到404
当第2页、第3页等没有返回任何结果时,将第2页重定向到/results/
的最佳方式是什么
我想写一个自定义的404视图来去掉URL参数,但是我想最好在列表视图中捕捉到这个
这是我的列表视图:
class ListingListView(ListView):
ORDER_BY_MAP = {
None: {
'directive': ListingsManager.CATEGORY_DEFAULT_ORDER,
},
'title': {
'label': 'Title A > Z',
'directive': ('title',),
},
'-title': {
'label': 'Title Z > A',
'directive': ('-title',),
},
'rating': {
'label': 'Rating',
'directive': ('-aggregate_rating', '-reviews_count'),
},
}
template_name = 'directory/listing_list.html'
context_object_name = 'listings'
paginate_by = 20
@cached_property
def country(self):
if 'country_slug' in self.kwargs:
country = get_object_or_404(Country, slug=self.kwargs['country_slug'])
set_user_language(self.request, country.language)
return country
else:
return None
@cached_property
def category(self):
if 'category_slug' in self.kwargs:
return get_object_or_404(Category.objects.select_related('parent').prefetch_related(
'children'), slug=self.kwargs['category_slug'])
else:
return None
@cached_property
def region(self):
if 'region_slug' in self.kwargs:
return get_object_or_404(Region, slug=self.kwargs['region_slug'])
else:
return None
def get_url_params(self):
return {k: v[0] for k, v in dict(self.request.GET).items()}
def get_queryset(self):
url_params = self.get_url_params()
user_order_by = url_params.get('order_by', None)
if user_order_by not in self.ORDER_BY_MAP:
user_order_by = None
country = self.country
category = self.category
region = self.region
queryset = Listing.objects.category_view(category, country, region)
return queryset.order_by(*self.ORDER_BY_MAP[user_order_by]['directive']).distinct()
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
country = self.country
region = self.region
listings = self.get_queryset()
context['category'] = category = self.category
if region:
context['title'] = f'{category.name} in {region.name}'
else:
context['title'] = category.name
# Add information for the category sidebar
context['up_links'] = []
if not category.is_root:
root_category = Category.objects.root()
context['up_links'].append({
'name': root_category.name,
'link': reverse('directory-category', args=[country.slug, root_category.slug])
})
if category.parent:
context['up_links'].append({
'name': category.parent.name,
'link': reverse('directory-category', args=[country.slug, category.parent.slug])
})
if region:
context['up_links'].append({
'name': category.name,
'link': reverse('directory-category', args=[country.slug, category.slug])
})
if category.has_regions:
context['region_links'] = {}
regions = Region.objects.filter(listings__categories=category, listings__countries=country).order_by('name')
for item in regions:
context['region_links'][item.name] = reverse('directory-region',
args=[country.slug, category.slug, item.slug])
context['subcategories'] = []
if category.is_root:
children = Category.objects.filter(parent__isnull=True).exclude(is_root=True)
else:
children = category.children
for c in children.order_by('nav_menu_order'):
context['subcategories'].append({
'name': c.name,
'thumbnail_image': c.thumbnail_image,
'link': reverse('directory-category', args=[country.slug, c.slug]),
})
# Do not show recommended badge in root category view
context['show_recommended_badge'] = not category.is_root
# Add sorting menu
context['order_by_URLs'] = {}
for key, value in self.ORDER_BY_MAP.items():
if key:
context['order_by_URLs'][value['label']] = f'{self.request.path}?order_by={key}'
# Add total listings count (template only counts listings on that page, not total)
context['listings_count'] = listings.count()
# Add sorting parameters so we can preserve sorting in pagination links
url_params = self.request.GET.copy()
url_params.pop('page', None)
context['url_params'] = url_params
return context
您可以通过
.objects.all().count()
查看模型的长度是否超过20(这是分页值)。从该值可以执行if/else语句并呈现视图。要么呈现页面本身/results/
,要么重定向到主页(或任何其他页面)。如果您可以发送视图的代码。py
试试看,在您的列表列表视图中,将分页方式=20
替换为列表分页方式=10
,这将覆盖默认分页方式
在get\u context\u data
方法中:
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
listings = Listing.objects.all() # change this according to your requirement
# paginate the listings
paginator = Paginator(listings, self.list_per_page)
page = self.request.GET.get('page')
try:
paged_listings = paginator.page(page)
except EmptyPage:
# in case if you got an empty page this will show the fist page
paged_listings = paginator.page(1)
context['paged_listings'] = paged_listings
return context
您必须导入类和异常
from django.core.paginator import Paginator, EmptyPage
注意:由于paged_清单
是上下文名称,您必须在模板中使用此上下文名称才能查看分页清单谢谢大家的回答
我最后做的是编写一个mixin,覆盖get
方法。这样,我就可以对所有分页的ListView重用它,而无需自定义分页器
这是混音器:
class SafePaginateMixin(object):
def get(self, *args, **kwargs):
try:
return super().get(*args, **kwargs)
except Http404:
page = self.request.GET.get('page', None)
if page:
return redirect(self.request.path)
else:
raise
然后将mixin添加到我的列表视图中
:
class ListingListView(SafePaginateMixin, ListView)
如果404由带有page
URL参数的URL引发,则重定向到请求路径。如果没有,则升起404