我有这个类用于创建和更新课程内容。如何从这个视图转换或编写API视图?
class ContentCreateUpdateView(TemplateResponseMixin, View):
module = None
model = None
obj = None
template_name = 'courses/manage/content/form.html'
def get_model(self, model_name):
if model_name in ['text', 'video', 'image', 'file']:
return apps.get_model(app_label='courses',
model_name=model_name)
return None
def get_form(self, model, *args, **kwargs):
Form = modelform_factory(model, exclude=['owner',
'order',
'created',
'updated'])
return Form(*args, **kwargs)
def dispatch(self, request, module_id, model_name, id=None):
self.module = get_object_or_404(Module,
id=module_id,
course__owner=request.user)
self.model = self.get_model(model_name)
if id:
self.obj = get_object_or_404(self.model,
id=id,
owner=request.user)
return super().dispatch(request, module_id, model_name, id)
def get(self, request, module_id, model_name, id=None):
form = self.get_form(self.model, instance=self.obj)
return self.render_to_response({'form': form,
'object': self.obj})
def post(self, request, module_id, model_name, id=None):
form = self.get_form(self.model,
instance=self.obj,
data=request.POST,
files=request.FILES)
if form.is_valid():
obj = form.save(commit=False)
obj.owner = request.user
obj.save()
if not id:
# new content
Content.objects.create(module=self.module,
item=obj)
return redirect('module_content_list', self.module.id)
return self.render_to_response({'form': form,
'object': self.obj})
我想要wite API代码使用Django restframework我建立LMS使用Django我支持和React在前端问题是如何写API,如果我有这样的硬代码
1条答案
按热度按时间8ftvxx2r1#
你可以写你说的所有方法。首先是删除方法。
删除:
列表: