Python 模板呈现期间出现Django错误,因为URL标记

Python 模板呈现期间出现Django错误,因为URL标记,python,django,python-2.7,url,model-view-controller,Python,Django,Python 2.7,Url,Model View Controller,错误: URL.py: from django.shortcuts import get_object_or_404, render from .models import College # Create your views here. def index(request): latest_college_list = College.objects.order_by('college_name') context = {'latest_college_list': lat

错误:

URL.py:

from django.shortcuts import get_object_or_404, render
from .models import College

# Create your views here.
def index(request):
    latest_college_list = College.objects.order_by('college_name')
    context = {'latest_college_list': latest_college_list}
    return render(request, 'app/index.html', context)

def detail(request, college_id):
    college = get_object_or_404(College, pk=college_id)
    return render(request, 'app/detail.html', {'college':college})
有人知道问题是什么吗?

问题是:

from django.conf.urls import include, url
from django.contrib import admin

urlpatterns = [
    url(r'^$', include('app.urls')),
    url(r'^login/$', include('login.urls')),
    url(r'^admin/', admin.site.urls),
]
应该是

url(r'^$', include('app.urls')),
注意,
$
应该被删除,因为它表示正则表达式模式的结束,并且它不会发现包含的url模式

其次,您需要在
include
中明确指定
名称空间。更多关于这方面的信息,请参阅


同样,在
登录/
URL模式匹配后删除
$

你的错误是什么?另外,您的根url配置是什么样子的?非常感谢!
from django.shortcuts import get_object_or_404, render
from .models import College

# Create your views here.
def index(request):
    latest_college_list = College.objects.order_by('college_name')
    context = {'latest_college_list': latest_college_list}
    return render(request, 'app/index.html', context)

def detail(request, college_id):
    college = get_object_or_404(College, pk=college_id)
    return render(request, 'app/detail.html', {'college':college})
from django.conf.urls import url
from . import views

app_name = "app"
urlpatterns = [
    # campusarchitecture.com/
    url(r'^$', views.index, name="index"),
    # /college_name
    url(r'^(?P<college_id>[0-9]+)/$', views.detail, name="detail")
]
from django.conf.urls import include, url
from django.contrib import admin

urlpatterns = [
    url(r'^$', include('app.urls')),
    url(r'^login/$', include('login.urls')),
    url(r'^admin/', admin.site.urls),
]
url(r'^$', include('app.urls')),
url(r'^', include('app.urls', namespace="app")),