Question

我最初只扩展了Group模型，虽然我让额外的字段正常工作，但当我切换回auth-user模型时，我最终会得到不一致的结果：

group1 = MyGroup(..)
user = group1.user_set.all()[0]
group2 = user.groups.all()[0]
group1 == group2 #(False!)
type(group1) #MyGroup..
type(group2) #Group..

我的下一个想法是扩展auth.User和auth.Group模型。

示例结构：

from django.contrib.auth.models import User As DjangoUser
from django.contrib.auth.models import Group As DjangoGroup

class MyGroup(DjangoGroup):
    extra1 = models.CharField(..)

class MyUser(DjangoUser):
    extra1 = models.CharField(..)

这可能吗？ I saw a bug-report stating it was fixed here.但是我没有看到实现这一目标的正确方法的示例，也不清楚错误修复后这部分是否必要：

manager = UserManager()
class MyUser(DjangoUser):
    signature = forms.CharField()
    post_count = forms.IntegerField()
    objects = manager
    _default_manager = manager

这种行为的任何想法/例子？我试图在用户模型中重新定义“组”链接，但这导致了验证错误。理想情况下，我希望能够运行上面的示例并使group1 == group2和type（group2）== type（MyGroup（））

Answer 1

您可以向django用户/组模型动态添加新字段，这样就不会创建新的类类型。有关参考，请参阅：http://code.google.com/p/comaie-django-groups/source/browse/trunk/src/comaie/django/groups/models.py?r=5

models.ForeignKey(
    Group,
    null            = True,
    blank           = True,
    related_name    = 'children',
    verbose_name    = _('parent'),
    help_text       = _('The group\'s parent group. None, if it is a root node.')
).contribute_to_class(Group, 'parent')

def get_all_groups(self):
    """
    Returns all groups the user is member of AND all parent groups of those
    groups.
    """
    direct_groups = self.groups.all()
    groups = set()

    for group in direct_groups:
        ancestors = group.get_ancestors().all()
        for anc in ancestors:
            groups.add(anc)
        groups.add(group)

    return groups

setattr(User, 'get_all_groups', get_all_groups)

在Django中扩展Auth用户和组模型

1 个答案: