如何扩展 Django Group 模型？

Question

有没有一种方法可以扩展内置的 Django Group 对象以添加附加属性，类似于扩展用户对象的方式？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文件中

AUTH_PROFILE_MODULE = 'app.UserProfile'

这让你：

profile = User.objects.get(id=1).get_profile()

是否有与此方法等效的扩展组的方法？如果没有，我可以采取其他方法吗？

Answer 1

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，并且管理站点不会选择任何新字段。

您需要向现有组中注入新字段：

if not hasattr(Group, 'parent'):
    field = models.ForeignKey(Group, blank=True, null=True, related_name='children')
    field.contribute_to_class(Group, 'parent')

要将方法添加到组中，子类化但将模型标记为代理：

class MyGroup(Group):

    class Meta:
        proxy = True

    def myFunction(self):
        return True

Answer 2

您可以创建一个子类化 Group 的模型，添加您自己的字段，并使用模型管理器返回您需要的任何自定义查询集。这是一个截断的示例，显示了我如何扩展组来代表与学校相关的家庭：

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

Answer 3

我设法通过@Semprini aswer 使用迁移。

所以我需要在我的团体相关字段中创建一个公司相关字段，所以在我的模型中我这样做了：

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，但第一个属于

auth

应用程序是在我的虚拟环境中创建的。文件看起来像这样：

# 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 迁移文件夹中创建的第二个如下所示：

# 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，解决方案是将我的 virtualenv 中创建的文件移至 myapp 迁移文件夹，然后再使用 makemigrations 生成另一个文件，但由于迁移应用于

auth

应用程序而不是

myapp

，我必须在中实现一种解决方法文件。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

变异和应用方法允许您从

auth

迁移迁移到

myapp

应用程序。

在第二个文件中，我只是更改依赖项以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()),
            ],
        ),
    ]

Answer 4

对我来说，工作解决方案基于：

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组做了什么：

首先，我创建了自己的 django 应用程序，命名为它

python Manage.py startapp auth_custom

代码部分：

在 auth_custom/models.py 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:

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 管理视图中得到这样的结果。

为了访问此自定义字段，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

如有错误请通知我，我会更正此答案。

Answer 5

您可以使用 django-group-model 包来覆盖组模型，就像处理用户模型一样。这是一个简短的例子。

安装包

pip install django-group-model

将其添加到顶部已安装的应用程序中。

INSTALLED_APPS = [
    ...
    'django_group_model',
    ...
]

安装后，我们将覆盖默认的组模型，将其称为角色，并添加类别字段作为自定义字段（可选）。

from django_group_model.models import AbstractGroup


class Role(AbstractGroup):
    category = models.CharField(max_length=50)

    class Meta:
        verbose_name = "role"
        verbose_name_plual = "roles"

设置

AUTH_GROUP_MODEL

设置。

AUTH_GROUP_MODEL = 'myapp.Role'

将组模型与您的用户模型一起使用

class User(AbstractUser, ...):
    ...
    roles = models.ManyToManyField(
        'myapp.Role',
        blank=True,
        related_name="user_set",
        related_query_name="user",
    )
    groups = None
    ...

不要忘记将默认值

groups

设置为

None

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