2012-07-25 60 views
8

当以下是错误Django的:错误调用元类基地

TypeError: Error when calling the metaclass bases metaclass conflict: the metaclass of a derived class must be a (non-strict) subclass of the metaclasses of all its bases

我的models.py

class Business(models.Model, forms.Form): 
    name = models.CharField(max_length=128) 
    tel_no = models.CharField(max_length=11) 
    address_ln1 = models.CharField(max_length=128) 
    address_ln2 = models.CharField(max_length=128) 
    city = models.CharField(max_length=64) 
    county = GBCountySelect() 
    postcode = GBPostcodeField() 
    website = models.URLField(max_length=128) 
# Logging Info 
    slug = models.SlugField() 
    date_added = models.DateField(auto_now_add=True) 
    time_added = models.TimeField() 
    added_by_user = models.CharField(max_length=64) 
    last_edit_time = models.TimeField(auto_now=True) 
    last_edit_date = models.DateField(auto_now=True) 

线我得到的误差范围内,有问题的类:

name = models.CharField(max_length=128) 

但我认为这意味着这一个:

class Business(models.Model, forms.Form): 

我不确定它到底意味着什么,我如何从同一个类中的models.Model和forms.Form继承我的模型?创建我的课程时,我不能传递两个值吗?如果是这样如何?

ANOTHER编辑

All my imports 
from django.db import models 
from django import forms 
from django.contrib.localflavor import generic 
from django.contrib.localflavor.gb.forms import GBPostcodeField, GBCountySelect 

完全回溯:

Traceback (most recent call last): 
    File "manage.py", line 10, in <module> 
    execute_from_command_line(sys.argv) 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/core/management/__init__.py", line 443, in execute_from_command_line 
    utility.execute() 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/core/management/__init__.py", line 382, in execute 
    self.fetch_command(subcommand).run_from_argv(self.argv) 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/core/management/base.py", line 196, in run_from_argv 
    self.execute(*args, **options.__dict__) 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/core/management/base.py", line 231, in execute 
    self.validate() 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/core/management/base.py", line 266, in validate 
    num_errors = get_validation_errors(s, app) 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/core/management/validation.py", line 30, in get_validation_errors 
    for (app_name, error) in get_app_errors().items(): 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/db/models/loading.py", line 158, in get_app_errors 
    self._populate() 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/db/models/loading.py", line 64, in _populate 
    self.load_app(app_name, True) 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/db/models/loading.py", line 88, in load_app 
    models = import_module('.models', app_name) 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/utils/importlib.py", line 35, in import_module 
    __import__(name) 
    File "/home/jws1000/envs/glutenfree/glutenfree/glutenfree/listings/models.py", line 9, in <module> 
    class Business(models.Model, forms.Form): 
    File "/home/jws1000/.virtualenvs/glutenfree/lib/python2.7/site-packages/django/db/models/base.py", line 41, in __new__ 
    new_class = super_new(cls, name, bases, {'__module__': module}) 
TypeError: Error when calling the metaclass bases 
    metaclass conflict: the metaclass of a derived class must be a (non-strict) subclass of the metaclasses of all its bases 
+0

你为什么想要?如果你想创建表单,请为此创建一个单独的类。 – Thomas 2012-07-25 19:54:00

+0

因为县和邮编与商业有关。将他们放在同一张桌子上不是明智的做法,因为我将逐一访问每条记录? – jdx 2012-07-25 19:57:11

+0

请全部显示错误。 http://sscce.org/ – Marcin 2012-07-25 20:01:16

回答

10

这就是问题所在:

class Business(models.Model, forms.Form): 

你试图从式样和继承。你不能,而你不应该。

您不能因为派生类的元类必须是其所有基类元类的(非严格)子类。形式所具有的元类:

__metaclass__ = DeclarativeFieldsMetaclass 

模式也有一个元类:

__metaclass__ = ModelBase 

如果你要做到这一点,你需要设置哪些来自那些派生元类。

但是,您不应该这样做,因为django具有ModelForms,它存在用于创建模型模型的表单,从而为您节省了复杂性的麻烦。只需停止从窗体继承。

+0

我删除了forms.Form,并使用本地风味材料保留了我的“表单”,并且它似乎可行。因为我不是从表格中调出任何东西。谢谢你的帮助。 – jdx 2012-07-25 20:14:32

+0

@jdx不客气。请接受此答案并投票。 – Marcin 2012-07-25 20:16:14