Django：调用元类库时出错

Posted 2023-02-23

【中文标题】Django：调用元类库时出错

【英文标题】：Django: Error when calling the metaclass bases

【发布时间】：2012-07-24 08:17:01

【问题描述】：

这是错误

TypeError：调用元类库时出错 元类冲突：派生类的元类必须是其所有基类的元类的（非严格）子类

models.py 中的相关类

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)

我收到错误的那一行：

name = models.CharField(max_length=128)

但我（认为）是指这个：

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

我不确定它的确切含义，我如何在同一个类中从 models.Model 和 forms.Form 继承我的模型？创建课程时我不能传递两个值吗？如果有怎么办？

另一个编辑

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

完整的追溯：

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

【问题讨论】：

您为什么要这样做？如果您想创建表单，请为此创建一个单独的类。

因为县和邮政编码与业务相关联。将它们放在同一张桌子上不是明智的吗，因为我将逐条访问每条记录？

请完整显示错误。 sscce.org

我已经编辑了我的原始帖子。任何线索将不胜感激，我认为这可能与我的进口中的冲突有关？

@jdx 我认为您需要了解 Django 表单是什么。它们与表格无关。

【参考方案1】：

这就是问题所在：

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

您正在尝试从 Model 和 Form 继承。你不能，也不应该。

你不能因为派生类的元类必须是其所有基类的元类的（非严格）子类。表单有一个元类：

__metaclass__ = DeclarativeFieldsMetaclass

模型也有一个元类：

__metaclass__ = ModelBase

如果您要这样做，您需要设置一个派生自这两者的元类。

但是，您不应该这样做，因为 django 有 ModelForms，它的存在是为了创建模型模型的表单，从而为您省去了这里复杂性的麻烦。停止从 Form 继承即可。

【讨论】：

我删除了 forms.Form，并使用本地风味的东西保留了我的“表单”，它似乎有效。因为我没有从 forms.* 调用中得到任何东西。感谢您的帮助。

@jdx 不客气。请接受这个答案并投票。