Question

如何将DetailView传递给网址中的'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。

我想改变下面的代码。

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'

如你所知，DetailView需要'slug'或'pk'......所以我必须将'slug'传递给DetailView ......但我不知道如何传递'slug'......

我正等着你的回答... ...

谢谢！

Answer 1

正确的方法应该是

return customerDetailView.as_view()(request, slug=slug)

如何在网址中的'slug'上传递DetailView？

1 个答案: