Question

有没有办法在基于功能的视图中访问模型实例？我在硬编码策略中尝试了下面的代码。我需要它是动态的。

形式：

<form method="POST" 
      action="{% url 'cadmin:toggle_status' model='Library' %}" #***** Model is hardcoded
      class="visible-lg-inline">
{% csrf_token %}
.....
</form>

网址：

path('toggle-status/<slug:model>', toggle_status, name='toggle_status'),

查看：

def toggle_status(request, model):
    /******* How can i access model instance here? *********/
    if request.POST:    
        toggle_status = request.POST.get('toggle-status')
    pk = request.POST.get('pk')
    if toggle_status and pk:
        if model == "Zone":
            Zone.objects.filter(pk=pk).update(status=toggle_status)
        if model == "Library":
            Library.objects.filter(pk=pk).update(status=toggle_status)
    return HttpResponseRedirect(reverse('cadmin:library_list'))

Answer 1

您没有模型实例。如果你想要一个，你需要得到它。

my_instance = MyModel.objects.get(pk=whatever)

Answer 2

像

这样的东西

return HttpResponseRedirect(reverse('cadmin:library_list', kwargs={'model': model}))

和

action="{% url 'cadmin:toggle_status' model={{model}} %}"

Answer 3

尝试以下
我使用eval来减少代码

def toggle_status(request, model):
    if request.POST:    
        toggle_status = request.POST.get('toggle-status')
        pk = request.POST.get('pk')
        if toggle_status and pk:
           if model == "Zone" or model == 'Library':
               # it's like I  was doing : Zone.Objects.get(pk=pk) or Library.object.get(pk=pk)
               instance = eval("%s.objects.get(pk=pk)" % model) # Not filter
               instance.status = toggle_status 
               instance.save()
           return HttpResponseRedirect(reverse('cadmin:library_list'))
  # rest of the code

在基于函数的视图中访问模型。

3 个答案: