我正在 django 中开发一个简单的“待办事项列表”应用程序(两个模型:列表和项目)。尝试学习和利用基于类的通用 View 。我有以下三个显示 View 正在运行，但我要求快速进行代码审查，看看在继续创建、更新和删除之前，我是否可以做些什么来提高对 django 的 ListView 和 DetailView 的使用/理解。预先感谢您的任何建议。
# class-based generic views for my lists, a list's items, and item detail...
class MyListsView(ListView):
"""Display My Lists"""
template_name = 'cmv_app/my_lists.html'
context_object_name = 'my_lists'
def get_queryset(self):
# override get_queryset() to filter in only lists belonging to the current user,
# eliminating the need to define a model in this class.
return List.objects.filter(user=self.request.user).order_by('-last_modified')
# override dispatch to decorate this view with login_required
@method_decorator(login_required)
def dispatch(self, *args, **kwargs):
return super(MyListsView, self).dispatch(*args, **kwargs)
class ListItemsView(ListView):
"""Display a list's items"""
template_name = 'cmv_app/list_items.html'
context_object_name = 'list_items'
def get_queryset(self):
# get a particular list's items
items = Item.objects.filter(list=self.kwargs['list_id']).order_by('-created_date')
return items
def get_context_data(self, **kwargs):
# Call the base implementation first to get a context
context = super(ListItemsView, self).get_context_data(**kwargs)
# Add the list, needed for the template to lookup urls for the item detail page
context['list'] = self.list
return context
@method_decorator(login_required)
def dispatch(self, request, *args, **kwargs):
self.list = get_object_or_404(List, pk=kwargs['list_id'])
# this is a security check: a user can only access his own lists
# (else raise PermissionDenied)
request = list_valid_for_user(request, self.list)
return super(ListItemsView, self).dispatch(request, *args, **kwargs)
class ItemDetailView(DetailView):
"""Show detail for one item"""
template_name = 'cmv_app/item_detail.html'
def get_object(self):
return self.item
@method_decorator(login_required)
def dispatch(self, request, *args, **kwargs):
# fetching list and item in here instead of in get_object() in order to implement
# security checks
self.list = get_object_or_404(List, pk=kwargs['list_id'])
self.item = get_object_or_404(Item, pk=kwargs['item_id'])
# user can only access his lists (else raise PermissionDenied)
request = list_valid_for_user(request, self.list)
# item must be associated with the specified list (else raise PermissionDenied)
request = item_valid_for_list(request, self.item, self.list)
return super(ItemDetailView, self).dispatch(request, *args, **kwargs)
def get_context_data(self, **kwargs):
# Call the base implementation first to get a context
context = super(ItemDetailView, self).get_context_data(**kwargs)
# Add in the elements needed for the template to construct url for link to list
context['list'] = self.list
return context
最佳答案
例如，对于 ListItemsViews，在您的调度中只需执行以下操作:
self.list = kwargs['list']
self.list_id = kwargs['list_id']
然后，转到get_context_data:
self.list = get_object_or_404(List, pk=self.list_id)
这将转到get_queryset:
items = Item.objects.filter(list=self.list_id).order_by('-created_date')
关于python - 简单的 django 基于类的通用 View : am I doing it right?，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/23877494/