modeladmin 中代理模型的用户权限

Question

使用代理模型时：

class Uf(models.Model):
...

class CustomUf(Uf):
    class Meta:
        proxy = True

class CustomUfAdmin(admin.ModelAdmin)

admin.site.register(CustomUf, CustomUfAdmin)

似乎只有超级用户才能通过管理站点访问 CustomUf...我不知道如何将 CustomUf 的权限授予普通用户...

Answer 1

好的，克里斯关于内容类型的评论给了我提示......

我在“admin.py”中定义代理对象时犯了一个错误。这样，您必须是超级管理员才能访问它。

如果我在models.py中定义了代理对象，那么内容类型就会出现并且一切正常...

Answer 2

您需要再次运行 syncdb 以便获取新的内容类型。

Answer 3

请参阅这个相关的 Django 问题：#11154

您可以通过手动将行添加到“auth_permission”表来克服这个问题，例如：

INSERT INTO "auth_permission" ("name","content_type_id","codename") 
    VALUES ('Can add proxy model name',{content_type_id},'add_proxy_model_name');

其中内容类型 id 是相关内容类型的整数 id。

Answer 4

我意识到这个问题不久前就结束了，但我正在分享对我有用的方法，以防它可能对其他人有所帮助。

事实证明，即使我创建的代理模型的权限列在父应用程序下，即使我授予我的非超级用户所有权限，它仍然被拒绝通过管理员访问我的代理模型。

如果您想避免 raw SQL 并在 Python 中编写修复脚本，您必须解决已知的 Django 错误 (https://code.djangoproject.com/ticket/11154) 并连接到 post_syncdb 信号以正确创建代理模型的权限。下面的代码是根据该线程上的一些 cmets 从 https://djangosnippets.org/snippets/2677/ 修改而来的。

我将它放在保存我的代理模型的 myapp/models.py 中。从理论上讲，这可以存在于django.contrib.contenttypes 之后的任何INSTALLED_APPS 中，因为它需要在为update_contenttypes 处理程序注册post_syncdb 信号之后加载，以便我们可以断开它。

def create_proxy_permissions(app, created_models, verbosity, **kwargs):
    """
    Creates permissions for proxy models which are not created automatically
    by 'django.contrib.auth.management.create_permissions'.
    See https://code.djangoproject.com/ticket/11154
    Source: https://djangosnippets.org/snippets/2677/

    Since we can't rely on 'get_for_model' we must fallback to
    'get_by_natural_key'. However, this method doesn't automatically create
    missing 'ContentType' so we must ensure all the models' 'ContentType's are
    created before running this method. We do so by un-registering the
    'update_contenttypes' 'post_syncdb' signal and calling it in here just
    before doing everything.
    """
    update_contenttypes(app, created_models, verbosity, **kwargs)
    app_models = models.get_models(app)
    # The permissions we're looking for as (content_type, (codename, name))
    searched_perms = list()
    # The codenames and ctypes that should exist.
    ctypes = set()
    for model in app_models:
        opts = model._meta
        if opts.proxy:
            # Can't use 'get_for_model' here since it doesn't return
            # the correct 'ContentType' for proxy models.
            # See https://code.djangoproject.com/ticket/17648
            app_label, model = opts.app_label, opts.object_name.lower()
            ctype = ContentType.objects.get_by_natural_key(app_label, model)
            ctypes.add(ctype)
            for perm in _get_all_permissions(opts, ctype):
                searched_perms.append((ctype, perm))

    # Find all the Permissions that have a content_type for a model we're
    # looking for. We don't need to check for codenames since we already have
    # a list of the ones we're going to create.
    all_perms = set(Permission.objects.filter(
        content_type__in=ctypes,
    ).values_list(
        "content_type", "codename"
    ))

    objs = [
        Permission(codename=codename, name=name, content_type=ctype)
        for ctype, (codename, name) in searched_perms
        if (ctype.pk, codename) not in all_perms
    ]
    Permission.objects.bulk_create(objs)
    if verbosity >= 2:
        for obj in objs:
            sys.stdout.write("Adding permission '%s'" % obj)


models.signals.post_syncdb.connect(create_proxy_permissions)
# See 'create_proxy_permissions' docstring to understand why we un-register
# this signal handler.
models.signals.post_syncdb.disconnect(update_contenttypes)