1

On Django I'm getting this traceback:

File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
  115.                         response = callback(request, *callback_args, **callback_kwargs)
File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/contrib/admin/options.py" in wrapper
  372.                 return self.admin_site.admin_view(view)(*args, **kwargs)
File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/utils/decorators.py" in _wrapped_view
  91.                     response = view_func(request, *args, **kwargs)
File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/views/decorators/cache.py" in _wrapped_view_func
  89.         response = view_func(request, *args, **kwargs)
File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/contrib/admin/sites.py" in inner
  202.             return view(request, *args, **kwargs)
File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/utils/decorators.py" in _wrapper
  25.             return bound_func(*args, **kwargs)
File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/utils/decorators.py" in _wrapped_view
  91.                     response = view_func(request, *args, **kwargs)
File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/utils/decorators.py" in bound_func
  21.                 return func(self, *args2, **kwargs2)
File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/db/transaction.py" in inner
  223.                 return func(*args, **kwargs)
File "/Users/eeytan/ddragon/lib/python2.7/site-packages/django/contrib/admin/options.py" in change_view
  1085.             form = ModelForm(request.POST, request.FILES, instance=obj)

Exception Type: TypeError at /admin/dragon_portal/parentprofile/1/
Exception Value: __init__() got multiple values for keyword argument 'instance'

For this code based on Creating one Django Form to save two models:

class ParentCreationForm(UserCreationForm):
    first_name  = forms.CharField(max_length=100)
    last_name   = forms.CharField(max_length=100)
    email       = forms.EmailField()
    #ice_contact = forms.CharField(max_length=100)
    #notes       = HTMLField()

    def __init__(self, instance=None, *args, **kwargs):
        _fields = ('username', 'first_name', 'last_name', 'email', 'password')
        _initial = model_to_dict(instance.dragonuser, _fields) \
            if instance is not None else {}
        kwargs['initial'] = _initial
        super(ParentCreationForm, self).__init__(instance=instance, *args, **kwargs)
        self.fields.update(fields_for_model(DragonUser, _fields))

As you can see, the signature of __init__() is not changed. Also, more strangely, Django's more detailed traceback shows that for each step in the traceback, the argv is always, {}, so I'm not even sure at what point of the traceback the error comes from.

4

2 回答 2

1

ModelForm的__init__函数签名不正确。

从BaseModelForm 的 Django 源代码中，您可以看到函数签名是：

def __init__(self, data=None, files=None, auto_id='id_%s', prefix=None,
                 initial=None, error_class=ErrorList, label_suffix=None,
                 empty_permitted=False, instance=None):

因此，如果有人像 Django 那样在第一个位置使用未命名参数实例化您的表单，那么您最终会遇到此错误。

我建议__init__像这样重写你的：

 def __init__(self, *args, **kwargs):
        instance = kwargs.get('instance')
        _fields = ('username', 'first_name', 'last_name', 'email', 'password')
        _initial = model_to_dict(instance.dragonuser, _fields) \
            if instance is not None else {}
        kwargs['initial'] = _initial
        super(ParentCreationForm, self).__init__(*args, **kwargs)
        self.fields.update(fields_for_model(DragonUser, _fields))
于 2013-10-30T02:29:03.727 回答
1

问题是 Python 不会让您在对 init 的调用中为“instance”指定关键字参数（就像在 .../django/contrib/admin/options.py:1085 中所做的那样），因为第一个参数由于参数的顺序，在调用中被映射到“实例”，然后第三个参数也被映射到“实例”，因为被指定为关键字参数。

要解决此问题，instance=None请从您的__init__签名中删除，而是在方法的第一行执行以下操作：

instance = kwargs.get("instance")

还要instance=instance从超级调用中删除，因为现在应该在 kwargs 中携带实例。

于 2013-10-30T02:29:11.043 回答