如何将DetailView传递给 url 中的“slug”？
首先，让我们看看我的代码。
网址.py
from django.conf.urls import patterns, include, url
urlpatterns = patterns('',
url(r'^customer/(?P<slug>[^/]+)/$', customerDetailView.as_view()),
)
视图.py
from django.views.generic import DetailView
class customerDetailView(DetailView):
context_object_name = 'customerDetail'
template_name = "customerDetail.html"
allow_empty = True
model = Customer
slug_field = 'name' # 'name' is field of Customer Model
现在，我的代码就像上面一样。
我想更改如下代码。
网址.py
from django.conf.urls import patterns, include, url
urlpatterns = patterns('',
url(r'^customer/(?P<slug>[^/]+)/$', customer),
)
视图.py
from django.views.generic import DetailView
class customerDetailView(DetailView):
context_object_name = 'customerDetail'
template_name = "customerDetail.html"
allow_empty = True
model = Customer
slug_field = 'name' # 'name' is field of Customer Model
def customer(request, slug):
if request.method == "DELETE":
pass # some code blah blah
elif request.method == "POST"
pass
elif request.method == "GET":
return customerDetailView.as_view(slug=slug)(request) # But this line is not working... just causing error TypeError, customerDetailView() received an invalid keyword 'slug'
如您所知，DetailView 需要“slug”或“pk”...所以我必须将“slug”交付给 DetailView...但我不知道如何交付“slug”...
我在显示器前等你的答案...
谢谢！