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我並不十分了解Django的評論框架中的get_comment_permalink。Django評論框架中的get_comment_permalink如何工作？
我使用Django的評論爲我的課Order創建了一些評論，默認情況下它顯示類似/comments/cr/18/1/#c1的url，並且該url永遠不存在。
我看了看評論的urls.py，它有這樣一行
urlpatterns += patterns('',
url(r'^cr/(\d+)/(.+)/$', 'django.contrib.contenttypes.views.shortcut', name='comments-url-redirect'),
)
它具有shortcut法views.py是
from django import http
from django.contrib.contenttypes.models import ContentType
from django.contrib.sites.models import Site, get_current_site
from django.core.exceptions import ObjectDoesNotExist
from django.utils.translation import ugettext as _
def shortcut(request, content_type_id, object_id):
"""
Redirect to an object's page based on a content-type ID and an object ID.
"""
# Look up the object, making sure it's got a get_absolute_url() function.
try:
content_type = ContentType.objects.get(pk=content_type_id)
if not content_type.model_class():
raise http.Http404(_(u"Content type %(ct_id)s object has no associated model") %
{'ct_id': content_type_id})
obj = content_type.get_object_for_this_type(pk=object_id)
except (ObjectDoesNotExist, ValueError):
raise http.Http404(_(u"Content type %(ct_id)s object %(obj_id)s doesn't exist") %
{'ct_id': content_type_id, 'obj_id': object_id})
try:
get_absolute_url = obj.get_absolute_url
except AttributeError:
raise http.Http404(_("%(ct_name)s objects don't have a get_absolute_url() method") %
{'ct_name': content_type.name})
absurl = get_absolute_url()
# Try to figure out the object's domain, so we can do a cross-site redirect
# if necessary.
# If the object actually defines a domain, we're done.
if absurl.startswith('http://') or absurl.startswith('https://'):
return http.HttpResponseRedirect(absurl)
# Otherwise, we need to introspect the object's relationships for a
# relation to the Site object
object_domain = None
if Site._meta.installed:
opts = obj._meta
# First, look for an many-to-many relationship to Site.
for field in opts.many_to_many:
if field.rel.to is Site:
try:
# Caveat: In the case of multiple related Sites, this just
# selects the *first* one, which is arbitrary.
object_domain = getattr(obj, field.name).all()[0].domain
except IndexError:
pass
if object_domain is not None:
break
# Next, look for a many-to-one relationship to Site.
if object_domain is None:
for field in obj._meta.fields:
if field.rel and field.rel.to is Site:
try:
object_domain = getattr(obj, field.name).domain
except Site.DoesNotExist:
pass
if object_domain is not None:
break
# Fall back to the current site (if possible).
if object_domain is None:
try:
object_domain = get_current_site(request).domain
except Site.DoesNotExist:
pass
# If all that malarkey found an object domain, use it. Otherwise, fall back
# to whatever get_absolute_url() returned.
if object_domain is not None:
protocol = request.is_secure() and 'https' or 'http'
return http.HttpResponseRedirect('%s://%s%s'
% (protocol, object_domain, absurl))
else:
return http.HttpResponseRedirect(absurl)
，它太複雜，我理解。
當Django說固定鏈接時，我考慮引用一個頁面上的特定位置（通常是標題）。例如，Django的評論框架文檔是鏈接＃1，並且您可以使用鏈接＃2固定「鏈接到評論」部分。
1. https://docs.djangoproject.com/en/dev/ref/contrib/comments/
2. https://docs.djangoproject.com/en/dev/ref/contrib/comments/#linking-to-comments
所以對於評論，不應該是相同的嗎？不應該只是簡單的#c1或沒有/comments/cr/18/1/...的東西？事實上，我甚至不知道從哪裏得到的Django和18 ... 1從shortcut方法，據我所知，18是content_type_id和1是object_id，但我怎麼能知道哪些類models.py是內容類型ID和對象ID？
不這似乎違背DRY pinciple /鬆耦合？我在'models.py'中定義了'def get_absolute_url（self）：return「/ doors/orders/{pk} /」.format（pk = self.pk）'並且定義了'r'^ orders /（？P我的門應用程序中的'urls.py'中的 \ d +）/ $''（完整路徑仍然是'/ doors/orders/1 /'之類的東西）。有沒有辦法「倒退」或「reverse_lazy」呢？ –
hobbes3
2012-03-27 20:27:21
這是另一個問題，所以我會簡短介紹一下，但是，你是對的！你應該做這樣的事情來獲取url：'from django.core import urlresolvers; urlresolvers.reverse（'doors_order_detail'，args = [self.pk]）'。當然，這需要您[命名您的訂單明細網址]（https://docs.djangoproject.com/en/dev/topics/http/urls/#id2）'doors_order_detail'，這在Django中非常標準。 – jpic 2012-03-27 20:32:05