如何将 DetailView 传递给 url 中的“slug”？
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【中文标题】如何将 DetailView 传递给 url 中的“slug”？【英文标题】：How to pass DetailView over the 'slug' in url? 【发布时间】：2013-04-03 02:30:00 【问题描述】：如何将DetailView 传递给 url 中的“slug”？
首先，让我们看看我的代码。
urls.py
from django.conf.urls import patterns, include, url
urlpatterns = patterns('',
url(r'^customer/(?P<slug>[^/]+)/$', customerDetailView.as_view()),
)
views.py
from django.views.generic import DetailView
class customerDetailView(DetailView):
context_object_name = 'customerDetail'
template_name = "customerDetail.html"
allow_empty = True
model = Customer
slug_field = 'name' # 'name' is field of Customer Model
现在，我的代码就像上面一样。
我想更改如下代码。
urls.py
from django.conf.urls import patterns, include, url
urlpatterns = patterns('',
url(r'^customer/(?P<slug>[^/]+)/$', customer),
)
views.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”传递给 DetailView...但我不知道如何提供“slug”...
我在监视器前等待你的回答...
谢谢！
【问题讨论】：
【参考方案1】：正确的方法应该是
return customerDetailView.as_view()(request, slug=slug)
【讨论】：
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