DJANGO与抽象基类投掷错误后添加后的规则

Question

将django-rules添加到我的django项目中，我遇到以下问题：我想在我的抽象基类中添加默认权限，并在需要时覆盖它们。在基类和继承类的示例下：

class BaseModel(RulesModelBaseMixin):
    company_id = models.ForeignKey('company.Company', on_delete=models.CASCADE)
    created_by = models.ForeignKey(
        'user.User', on_delete=models.SET_NULL, null=True, blank=True)

    class Meta:
        abstract = True
        rules_permissions = {
            "can_create": can_create_in_company | is_superuser,
            "can_view": can_view_in_company | is_author | is_superuser
        }

class Ticket(RulesModelMixin, BaseModel, metaclass=RulesModelBase):
    name = models.CharField(max_length=512, null=True, blank=True)

添加此抽象基类后，有一个看似无关的错误消息：

Traceback (most recent call last):
    File "REPO_PATH/.venv/lib/python3.10/site-packages/django/utils/module_loading.py", line 30, in import_string
    return cached_import(module_path, class_name)
    File "REPO_PATH/.venv/lib/python3.10/site-packages/django/utils/module_loading.py", line 15, in cached_import
    import_module(module_path)
    File "/opt/homebrew/Cellar/[email protected]/3.10.4/Frameworks/Python.framework/Versions/3.10/lib/python3.10/importlib/__init__.py", line 126, in import_module
    return _bootstrap._gcd_import(name[level:], package, level)
    File "<frozen importlib._bootstrap>", line 1050, in _gcd_import
    File "<frozen importlib._bootstrap>", line 1027, in _find_and_load
    File "<frozen importlib._bootstrap>", line 1006, in _find_and_load_unlocked
    File "<frozen importlib._bootstrap>", line 688, in _load_unlocked
    File "<frozen importlib._bootstrap_external>", line 883, in exec_module
    File "<frozen importlib._bootstrap>", line 241, in _call_with_frames_removed
    File "REPO_PATH/app/api/pagination.py", line 10, in <module>
    from ticket.models import Ticket, TicketStatus, TicketType
    File "REPO_PATH/app/ticket/models.py", line 106, in <module>
    class Ticket(RulesModelMixin, BaseModel, metaclass=RulesModelBase):
    File "REPO_PATH/.venv/lib/python3.10/site-packages/rules/contrib/models.py", line 36, in __new__
    new_class._meta.rules_permissions = perms
AttributeError: type object 'Ticket' has no attribute '_meta'. Did you mean: 'Meta'?

它似乎与以后的自定义分页有关，但我不认为这是问题的原因，就像以前起作用一样。甚至在

Answer 1

1. basemodel不应子类rulesmodelbasemixin，该用作metaclass。 basemodel不需要metaclass = rulesmodelbasemixin，因为ticket has metaClass = pruremodelbase和prure modelbase子类RulesModelBasemixin。
2. basemodel应该子类模型，因为它定义了要继承的模型字段。 模型还提供了您可能期望的许多功能。
3. Ticket需要为具有Metaclass = ModelBase（例如model）的类子类次级类在问题中显示）。由于票证子类basemodel，我们将更改为子类model ，它无需明确的子类Model> Model 。
4. django-rules不支持Abstract Model Meta中的定义rules_permissions。您可以实现preprocess_rules_permissions类方法，以在子类上动态定义该方法。您还需要将baseModel放在pruremodelmixin ticket 中以覆盖该类方法。

# class BaseModel(RulesModelBaseMixin):
class BaseModel(Model):
    ...

    class Meta:
        abstract = True
        # rules_permissions = {
        #     "can_create": can_create_in_company | is_superuser,
        #     "can_view": can_view_in_company | is_author | is_superuser
        # }

    @classmethod
    def preprocess_rules_permissions(cls, perms):
        perms.update({
            "can_create": can_create_in_company | is_superuser,
            "can_view": can_view_in_company | is_author | is_superuser,
        })


# class Ticket(RulesModelMixin, BaseModel, metaclass=RulesModelBase):
class Ticket(BaseModel, RulesModelMixin, metaclass=RulesModelBase):
    ...

您可以通过制作basemodel子类rulesmodel：

class BaseModel(RulesModel):
    ...

    class Meta:
        abstract = True

    @classmethod
    def preprocess_rules_permissions(cls, perms):
        perms.update({
            "can_create": can_create_in_company | is_superuser,
            "can_view": can_view_in_company | is_author | is_superuser,
        })


class Ticket(BaseModel):
    ...