我有两个视图,其定义如下:电话DJANGO视图从另一视图返回结果
class ListResultView(LoginRequiredMixin, ListView):
model = Result
class GalleryView(LoginRequiredMixin, ListView):
model = Result
template = 'gallery.html'
context_object_name = 'gallery'
所以ListResultView()使用隐式定义的result_list.html作为模板和结果impliciyly定义为上下文模型,而GalleryView(它是相同数据的更漂亮的列表)使用明确定义的模板'gallery.html',上下文对象被定义为'画廊'。
我打电话给他们使用以下urls.py(这是主要的urls.py,而不是包括一个):
from django.conf.urls import patterns, include, url
import lc.views
from django.views.generic import TemplateView
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
url(r'^$', TemplateView.as_view(template_name="about.html"),
name='indexpage'),
url(r'^gallery/$', lc.views.GalleryView.as_view(), name='gallery'),
url(r'^admin/', include(admin.site.urls), name='admin'),
url(r'^listquery/', lc.views.ListCView.as_view(),
name='s_queries',),
url(r'^listresult/', lc.views.ListResultView.as_view(),
name='s_results',),
url(r'^new/', lc.views.CreateCQuery.as_view(),
name='query_new',),
url(r'^login/$', 'django.contrib.auth.views.login', name='login'),
url(r'^logout/$', 'django.contrib.auth.views.logout', {'next_page': '/login'}, name='logout'),
url(r'^deletes/(?P<pk>\d+)/$', lc.views.DeleteS.as_view(), name='delete_s'),
url(r'^deleter/(?P<pk>\d+)/$', lc.views.DeleteResult.as_view(), name='delete_result'),
url(r'^resultview/(?P<pk>\d+)/$', lc.views.ResultDetailView.as_view(),
name='resultview'),
url(r'^notyet', TemplateView.as_view(template_name="not_impl.html"), name="notyet",),
)
是打我的问题是,当我叫URL图库视图,我得到ListResultView响应。我看不到任何错误消息,并怀疑是否有人可以指出我要出错的位置,或者如何调试它。我目前的想法是丢弃基于类的视图并将其重写为基于函数的视图,这样我就可以更好地处理发生的事情,但是由于时间压力,我宁愿不这样做。
试试'url(r'^ gallery/$'' – pynovice
你可以显示完整的urls.py吗?是通过include加载的文件 - 如果是的话,你能显示父文件吗? –
没有变化,谢谢认为虽然! – arashiyama