键入对象' pizzamenu'没有属性' _default_manager'
我尝试在模板上访问pizzadetailview时有以下错误:attributeError at/pizza/6/ 类型对象“ pizzamenu”没有属性'_default_manager'。
问题在哪里?
models.py
class PizzaMenu(models.Model):
name = models.CharField(max_length=30)
description = models.TextField()
ingredients = models.CharField(max_length=100)
price = models.DecimalField(max_digits=4, decimal_places=2)
def __str__(self):
return self.name
class Meta:
ordering = ["price"]
views.py
from django.views.generic import TemplateView, ListView, DetailView
from .models import PizzaMenu
class IndexView(TemplateView):
template_name = "index.html"
class PizzaMenu(ListView):
model = PizzaMenu
template_name = "menu.html"
context_object_name = "pizza_menu"
# ordering = ["price"]
class PizzaDetailView(DetailView):
model = PizzaMenu
template_name = "pizza_detail.html"
context_object_name = "pizza_detail"
class About(TemplateView):
template_name = "about.html"
urls.py
from pizza.views import IndexView, PizzaMenu, About, PizzaDetailView
urlpatterns = [
path("", IndexView.as_view(), name="home"),
path("menu/", PizzaMenu.as_view(), name="menu"),
path("about/", About.as_view(), name="about"),
path("pizza/<int:pk>/", PizzaDetailView.as_view(), name="pizza_detail"),
path("admin/", admin.site.urls),
]
I have following error while i try to reach PizzaDetailView on template: AttributeError at /pizza/6/
type object 'PizzaMenu' has no attribute '_default_manager'.
Where is the problem?
models.py
class PizzaMenu(models.Model):
name = models.CharField(max_length=30)
description = models.TextField()
ingredients = models.CharField(max_length=100)
price = models.DecimalField(max_digits=4, decimal_places=2)
def __str__(self):
return self.name
class Meta:
ordering = ["price"]
views.py
from django.views.generic import TemplateView, ListView, DetailView
from .models import PizzaMenu
class IndexView(TemplateView):
template_name = "index.html"
class PizzaMenu(ListView):
model = PizzaMenu
template_name = "menu.html"
context_object_name = "pizza_menu"
# ordering = ["price"]
class PizzaDetailView(DetailView):
model = PizzaMenu
template_name = "pizza_detail.html"
context_object_name = "pizza_detail"
class About(TemplateView):
template_name = "about.html"
urls.py
from pizza.views import IndexView, PizzaMenu, About, PizzaDetailView
urlpatterns = [
path("", IndexView.as_view(), name="home"),
path("menu/", PizzaMenu.as_view(), name="menu"),
path("about/", About.as_view(), name="about"),
path("pizza/<int:pk>/", PizzaDetailView.as_view(), name="pizza_detail"),
path("admin/", admin.site.urls),
]
如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱,如果其他用户或者作者回复了您的评论,将不能在第一时间通知您!
发布评论
评论(1)
不要命名您的视图
pizzamenu
,它将覆盖对其他视图的pizzamenu
的引用。通常,基于班级的视图具有…查看
后缀,因此:Don't name your view
PizzaMenu
, it will override the reference to thePizzaMenu
for the other views. Usually class-based views have a…View
suffix, so: