Django的功能以基於類視圖

Question

我有此URL是http://127.0.0.1:8000/upload/picturelist/1，這使得USER_ID = 1，

以我urls.py

url(r'^picturelist/(?P<user_id>\d+)$', views.pictureList), 

以我view.py

def pictureList(request, user_id): 
    if int(user_id) != request.user.id: 
     raise PermissionDenied 

如何使這個基於功能的視圖使用createview？

class pictureList(CreateView): 

Answer 1

你可以做這樣的事情：

在urls.py：url(r'^picturelist/(?P<user_id>\d+)$', views.MakeItView.as_view()),

在views.py：

class MakeItView(CreateView): 
    model = myModel 
    template_name = 'whatever.html' 

    def get_context_data(self, **kwargs): 
     context = super(MakeItView, self).get_context_data(**kwargs) 
     if int(self.kwargs['user_id']) != self.request.user.id: 
      raise PermissionDenied 
     return context 

Answer 2

我從來沒有用過CreateView，但這裏是我收集來自閱讀文檔：

你可以通過定義form_valid做到這一點：

觀點：

class pictureList(CreateView): 
    model = YourModelHere 
    fields = ['whatever','fields','you','want','edited'] 

    def form_valid(self, form): 
    record = form.save(commit = False) 

    # assuming the user id is associated 
    # to the model with fieldname user_id 

    if (self.request.user == record.user_id): 
     record.save() 
     return HttpResponseRedirect(self.get_success_url()) 
    # not sure if this works: 
    return self.form_invalid() 

然後模板將在'yourappname/yourmodelhere_form.html'。

查看CreateView舉例。