我应该如何在django中制作详细信息,给予404
在文件“ views.py”中,我添加了下一个代码:
from django.views.generic import DetailView
class NewsDetailView(DetailView):
model = Articles
template_name = "news/details_view.html"
context_object_name = 'article'
然后在文件urls.py中添加到urlpatterns下一个:
path('<int:pk>', views.NewsDetailView.as_view(), name="news_detail"),
并创建了模板“ lotess_view.html”。 当我写信给浏览器时:
in file "views.py" i added next code:
from django.views.generic import DetailView
class NewsDetailView(DetailView):
model = Articles
template_name = "news/details_view.html"
context_object_name = 'article'
then in the file urls.py I added to urlpatterns next:
path('<int:pk>', views.NewsDetailView.as_view(), name="news_detail"),
and created the template "details_view.html".
When i write to browser:
http://127.0.0.1:8000/news/1
it gives 404 error
404 error in DetailView
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转到< admin站点,并添加至少一个
articles
对象。另外,您可以在Shell中创建它:
Go to Admin Site and add at least one
Articles
object.Alternatively, you can create it in shell: