我有以下代码,我希望form_invalid
方法返回与success_url
. 我一直在考虑子分类CreateView
,但我想知道公众舆论。如何实现上述的事情?
class ProgramNew(CreateView):
form_class = ProgramForm
template_name = 'programs/program_list.html'
success_url = '/manage/programs'
....
....
....
def form_invalid(self, form):
# How to return to self.success_url?
return super(ProgramNew, self).form_invalid(form)
苏丹