__init__() takes exactly 1 argument (7 given)

Question

What Is lacking in my implementation with this error Model: ``` class Person(models.Model): last_name = models.CharField(max_length=256) first_name = models.CharField(max_length=256) class Meta: abstract = True class Supplier(Person): def __init__(self): super(Supplier, self).__init__() class Item(models.Model): name = models.CharField(max_length=256) supplier = models.ForeignKey(Supplier) ... def __unicode__(self): return self.name ``` Views: ``` def ItemNew(request): if request.method == "POST": post_item = ItemNewForm(request.POST) ... else: item_form = ItemNewForm() return render(request, "item_new.html", { 'item_form' : item_form, }) ``` Forms: ``` class ItemNewForm(forms.ModelForm): class Meta: model = Item ``` Html: ``` ... ... ``` Backtrace: ``` TypeErro…

Accepted Answer

The problem is in

```
class Supplier(Person):

def __init__(self):
        super(Supplier, self).__init__()
```

Have a look at django.db.models.Model source code

```
class Model(six.with_metaclass(ModelBase)):
    _deferred = False

def __init__(self, *args, **kwargs):
```

Your `__init__` function get only the instance itself as argument, but django is passing probably more arguments. That's why you need to use `*args, **kwargs`.

To better understand `*args` and `**kwargs` you can have a look at this

init() takes exactly 1 argument (7 given)

Solution

Problem