抽象表单Django：__ init __（）获得了多个参数值

Question

我想创建一个AbstractEditForm（继承自ModelForm）表单，从中可以有多个表单从中继承。但是我收到了这个错误：

这是我的forms.py

# This is the abstract form that I want to inherit other forms from
class AbstractEditForms(forms.ModelForm):

    def __init__(self, id_fields=None, ref_field=None,
             model=None, *args, **kwargs):

        self.id_fields = id_fields
        self.changed_fields = {}
        self.ref_field = ref_field
        self.model_ = model
        self.ref_id_changed = False
        self.check_ref_id()

    try:
        if 'id_fields' in kwargs.keys():
            del kwargs['id_fields']
        if 'ref_fields' in kwargs.keys():
            del kwargs['ref_fields']
    except KeyError as e:
        print('Error in AbstractionEditForms :', str(e))

    super(AbstractEditForms, self).__init__(*args, **kwargs)



# This is the form that I want to use
class SchemeEditForm(AbstractEditForms):

    class Meta:
        model = SchemeModel
        exclude = ['created_on', 'financial_year']
        widgets = {
            'as_ref_id': forms.TextInput(attrs={'readonly': 'readonly',
                                                'placeholder': 'Auto Generated '
                                            }),
            'admin_sanction_amount': forms.HiddenInput(),
            'updated_on': forms.HiddenInput(),
    }

views.py：

def edit_scheme_form_view(request, pk=None):
    assert pk is not None, 'pk cannot be None, scheme edit form'
    instance = get_object_or_404(SchemeModel, pk=pk)
    id_fields = ['technical_authority', 'dept_name', ]
    model = SchemeModel
    ref_field = "as_ref_id"

    if request.method == 'GET':
        scheme_form = SchemeEditForm(None, id_fields=id_fields, ref_field="as_ref_id",
                                     model=model, instance=instance)
        context = {
            'form': scheme_form
        }
        return render(request, 'Forms/forms/SchemeForm.html', context=context)

    if request.method == 'POST':
        scheme_form = SchemeEditForm(request.POST, id_fields=id_fields, ref_field="as_ref_id",
                                     model=SchemeModel, instance=instance)

        if scheme_form.is_valid():

            instance = scheme_form.save()
            return generate_success_page(request, instance,"Scheme Edit Success",
                                         "Scheme Details - Edited", nav_dict=None,
                                         util_dict=None)
        else:
            return render(request, 'Forms/forms/SchemeForm.html', {'form': scheme_form})

错误traceback：

Internal Server Error: /edit/admin-sanction-form/1/
Traceback (most recent call last):
  File "C:\Python35\lib\site-packages\django\core\handlers\base.py", line 149, in get_response
response = self.process_exception_by_middleware(e, request)
  File "C:\Python35\lib\site-packages\django\core\handlers\base.py", line 147, in get_response
response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "C:\Users\karth\PycharmProjects\phc\edit_forms\views.py", line 17, in edit_scheme_form_view
model=model, instance=instance
TypeError: __init__() got multiple values for argument 'id_fields'
[04/Sep/2016 20:41:08] "GET /edit/admin-sanction-form/1/ HTTP/1.1" 500 63681

我知道这是初始化形式的问题，无论是AbstractEditForm还是SchemeEditForm。

非常感谢任何帮助。

Answer 1

SchemeEditForm正在将您传递的第一个位置参数解释为id_fields参数。当您稍后尝试按名称传递id_fields时，它认为它正在获得该参数的副本并且您收到错误。

尝试更改__init__()方法，以便在关键字参数之前接受任意位置参数：

def __init__(self, *args, id_fields=None, ref_field=None, model=None, **kwargs):

我不确定这是否会在您的表单功能方面为您提供所需的结果，但它会消除您所看到的错误。 请注意，此方法不在Python 2.X 中工作，只有Python 3。