2011-12-18 74 views
1

G'day那里。我现在比任何事情都有点烦恼。正如预期的,下面的代码完全可以工作。基本上它是一个ModelForm,它动态地将它的模型基于url中接收的字符串,或者基于实例的类(如果提供的话)。将Django Form类变量传递给嵌套类Meta,如何?

我的问题是,是否可以通过传递model_name变量将其抽象到另一个模块forms.py中。我可以将model_name传递给类没有问题,但我不知道如何将它传递给Meta后。有没有简单的方法来做到这一点?如果不这样做,但它会让我的代码堆更加整洁。

@user_passes_test(lambda u: u.is_staff, login_url="%slogin/" % NINJA_ADMIN_URL_PREFIX) 
def content_form(request, model_name=None, edit=False, call_name=''): 
    if edit: 
     content = Content.objects.get(call_name=call_name) 
     model_name = content.fields.__class__.__name__ 

    class ContentForm(forms.ModelForm): 
     parent = ModelTextField(queryset=Content.objects.all(), widget=JQueryAutocomplete(
         source_url='%sjson/call_names.json' % NINJA_ADMIN_URL_PREFIX, jquery_opts = {'minLength': 2}, 
         override_label='item.fields.call_name', override_value='item.fields.call_name'), 
         required=False) 

     class Meta(): 
      model = get_ninja_type(model_name) 
      widgets = { 
        'ninja_type': forms.widgets.HiddenInput(), 
       } 

     def clean_parent(self): 
      call_name = self.cleaned_data['parent'] 
      if call_name: 
       try: 
        parent = Content.objects.get(call_name=call_name) 
       except Content.DoesNotExist: 
        raise forms.ValidationError("The call name '%s' doesn't exist. Choose another parent." % call_name) 
       return parent 
      else: 
       return None 

    if request.method == 'POST': 
     if edit: 
      form = ContentForm(request.POST, instance=content.fields) 
     else: 
      form = ContentForm(request.POST) 
     if form.is_valid(): 
      content = form.save() 
      messages.success(request, "Your new content has been saved.") 
      return HttpResponseRedirect('%scontent/' % NINJA_ADMIN_URL_PREFIX) 

    else: 
     if edit: 
      form = ContentForm(instance=content.fields) 
     else: 
      form = ContentForm(initial={'ninja_type': model_name.lower(), 
            'author': request.user}) 
    if edit: 
     page_title = 'Edit %s' % model_name 
    else: 
     page_title = 'Create New %s' % model_name 

    return render(request, 'ninja/admin/content_form.html', { 
        'form': form, 
        'ninja_type': model_name, 
        'page_title': page_title, 
        'edit': edit, 
        'meta_field_names': NINJA_META_FIELD_NAMES, 
        }) 
+0

我有一个想法,也许表单可能被封装在一个函数返回一个类,然后调用该函数。当我有机会时会尝试。 – MalucoMarinero 2011-12-18 05:00:14

+0

是的,一个返回类的函数工作正常。一旦网站让我回复,我会发布答案的来源。 (声誉太低) – MalucoMarinero 2011-12-18 05:42:23

回答

1

当我想到这个问题时,答案很明显。返回一个类的函数工作正常。这是views.py的开始部分。我认为这可能是做到这一点的最佳方式,或者至少是最简洁的方式。

@user_passes_test(lambda u: u.is_staff, login_url="%slogin/" % NINJA_ADMIN_URL_PREFIX) 
def content_form(request, model_name=None, edit=False, call_name=''): 
    if edit: 
     content = Content.objects.get(call_name=call_name) 
     model_name = content.fields.__class__.__name__ 

    ContentForm = get_content_form(model_name) 

    if request.method == 'POST': 
     if edit: 
      form = ContentForm(request.POST, instance=content.fields) 
     else: 

这是forms.py条目。

def get_content_form(model_name): 
    class DynamicContentForm(forms.ModelForm): 
     parent = ModelTextField(queryset=Content.objects.all(), widget=JQueryAutocomplete(
         source_url='%sjson/call_names.json' % NINJA_ADMIN_URL_PREFIX, jquery_opts = {'minLength': 2}, 
         override_label='item.fields.call_name', override_value='item.fields.call_name'), 
         required=False) 

     class Meta(): 
      model = get_ninja_type(model_name) 
      widgets = { 
        'ninja_type': forms.widgets.HiddenInput(), 
       } 

     def clean_parent(self): 
      call_name = self.cleaned_data['parent'] 
      if call_name: 
       try: 
        parent = Content.objects.get(call_name=call_name) 
       except Content.DoesNotExist: 
        raise forms.ValidationError("The call name '%s' doesn't exist. Choose another parent." % call_name) 
       return parent 
      else: 
       return None 
    return DynamicContentForm