有没有办法扩展内置的Django Group对象以添加类似于扩展用户对象的方式的其他属性？使用用户对象，您可以执行以下操作：
class UserProfile(models.Model):
user = models.OneToOneField(User)
并将以下内容添加到settings.py文件
AUTH_PROFILE_MODULE = 'app.UserProfile'
让你：
profile = User.objects.get(id=1).get_profile()
是否有任何相当于扩展群组的方法？如果没有，我可以选择其他方法吗？
答案 0 :(得分：43)
如果您只是对Group对象进行子类化，那么默认情况下它将创建一个新的数据库表，而admin网站将不会获取任何新字段。
您需要将新字段注入现有组：
if not hasattr(Group, 'parent'):
field = models.ForeignKey(Group, blank=True, null=True, related_name='children')
field.contribute_to_class(Group, 'parent')
要向Group添加方法，请将模型标记为代理：
class MyGroup(Group):
class Meta:
proxy = True
def myFunction(self):
return True
答案 1 :(得分：23)
您可以创建一个模型，该模型包括Group，添加您自己的字段，并使用Model Manager返回您需要的任何自定义查询集。这是一个截断的示例，显示了我如何将Group扩展为代表与学校相关联的家庭：
from django.contrib.auth.models import Group, User
class FamilyManager(models.Manager):
"""
Lets us do querysets limited to families that have
currently enrolled students, e.g.:
Family.has_students.all()
"""
def get_query_set(self):
return super(FamilyManager, self).get_query_set().filter(student__enrolled=True).distinct()
class Family(Group):
notes = models.TextField(blank=True)
# Two managers for this model - the first is default
# (so all families appear in the admin).
# The second is only invoked when we call
# Family.has_students.all()
objects = models.Manager()
has_students = FamilyManager()
class Meta:
verbose_name_plural = "Families"
ordering = ['name']
def __unicode__(self):
return self.name
答案 2 :(得分：4)
对我来说，基于以下解决方案：
https://docs.djangoproject.com/pl/1.11/topics/auth/customizing/#extending-user
让我解释一下我对使用电子邮件别名扩展默认模型的网上论坛所做的操作：
首先，我创建了自己的django应用程序
python manage.py startapp auth_custom
代码部分：
在 auth_custom / models.py 中，我创建了对象 CustomGroup
from django.contrib.auth.models import Group
from django.db import models
class CustomGroup(models.Model):
"""
Overwrites original Django Group.
"""
def __str__(self):
return "{}".format(self.group.name)
group = models.OneToOneField('auth.Group', unique=True)
email_alias = models.EmailField(max_length=70, blank=True, default="")
在 auth_custom / admin.py 中：
from django.contrib.auth.admin import GroupAdmin as BaseGroupAdmin
from django.contrib.auth.models import Group
class GroupInline(admin.StackedInline):
model = CustomGroup
can_delete = False
verbose_name_plural = 'custom groups'
class GroupAdmin(BaseGroupAdmin):
inlines = (GroupInline, )
# Re-register GroupAdmin
admin.site.unregister(Group)
admin.site.register(Group, GroupAdmin)
进行迁移后，我在Django Admin视图中得到了这样的结果。
要访问此自定义字段，您必须输入：
from django.contrib.auth.models import Group
group = Group.objects.get(name="Admins") # example name
email_alias = group.customgroup.email_alias
如果有任何错误，请通知我，我会纠正此答复。
答案 3 :(得分：0)
我设法通过@Semprini aswer使用迁移。
所以我需要在我的组相关领域中创建一个公司相关领域，所以在我的模型中我这样做：
if not hasattr(Group, 'company'):
field = models.ForeignKey(Company, on_delete=models.DO_NOTHING, null=True)
field.contribute_to_class(Group, 'company')
class Group(Group):
class Meta:
proxy = True
然后我运行manage.py makemigrations。这创建了2个文件。一个依赖于另一个，但是第一个属于auth应用的是在我的虚拟环境中创建的。这些文件如下所示：
# Generated by Django 2.2.5 on 2019-10-08 16:00
from django.db import migrations, models
import django.db.models.deletion
class Migration(migrations.Migration):
dependencies = [
('myapp', '0013_guestuser_permissions_20190919_1715'),
('auth', '0011_update_proxy_permissions'),
]
operations = [
migrations.AddField(
model_name='group',
name='company',
field=models.ForeignKey(
null=True, on_delete=django.db.models.deletion.DO_NOTHING, to='myapp.Company'),
),
]
在myapp migrations文件夹中创建的第二个文件如下：
# Generated by Django 2.2.5 on 2019-10-08 16:00
import django.contrib.auth.models
from django.db import migrations
class Migration(migrations.Migration):
dependencies = [
('auth', '0012_group_company_20191008'),
('myapp', '0013_guestuser_permissions_20190919_1715'),
]
operations = [
migrations.CreateModel(
name='Group',
fields=[
],
options={
'proxy': True,
'indexes': [],
'constraints': [],
},
bases=('auth.group',),
managers=[
('objects', django.contrib.auth.models.GroupManager()),
],
),
]
因此，解决方案是将我在virtualenv中创建的文件移动到myapp migrations文件夹中，而不是另一个使用makemigrations生成的文件中，但是由于迁移是应用于auth应用而不是myapp必须在文件中实施解决方法。所以现在的最终文件是：
# Generated by Django 2.2.5 on 2019-10-08 16:00
from django.db import migrations, models
import django.db.models.deletion
class Migration(migrations.Migration):
dependencies = [
('myapp', '0013_guestuser_permissions_20190919_1715'),
('auth', '0011_update_proxy_permissions'),
]
operations = [
migrations.AddField(
model_name='group',
name='company',
field=models.ForeignKey(
null=True, on_delete=django.db.models.deletion.DO_NOTHING, to='myapp.Company'),
),
]
def mutate_state(self, project_state, preserve=True):
"""
This is a workaround that allows to store ``auth``
migration outside the directory it should be stored.
"""
app_label = self.app_label
self.app_label = 'auth'
state = super(Migration, self).mutate_state(project_state, preserve)
self.app_label = app_label
return state
def apply(self, project_state, schema_editor, collect_sql=False):
"""
Same workaround as described in ``mutate_state`` method.
"""
app_label = self.app_label
self.app_label = 'auth'
state = super(Migration, self).apply(project_state, schema_editor, collect_sql)
self.app_label = app_label
return state
mutate应用方法允许您从auth迁移到myapp应用。
在第二个文件中，我只是将依赖关系更改为依赖于新创建的文件：
# Generated by Django 2.2.5 on 2019-10-08 16:00
import django.contrib.auth.models
from django.db import migrations
class Migration(migrations.Migration):
dependencies = [
('myapp', '0014_group_company_20191008'),
('myapp', '0013_guestuser_permissions_20190919_1715'),
]
operations = [
migrations.CreateModel(
name='Group',
fields=[
],
options={
'proxy': True,
'indexes': [],
'constraints': [],
},
bases=('auth.group',),
managers=[
('objects', django.contrib.auth.models.GroupManager()),
],
),
]