我对 ORM 很陌生，所以我需要一些帮助来为我的 Flask 应用程序创建菜单。我已经实现了模型 MenuItem（src 在下面）。
要点是：
- 它与自身具有一对多的关系。
- 它与模型具有多对多关系
Role
我的问题：
我可以跳过删除 generate_menu fn 中不可访问的子对象吗？这段代码（递归方法mark_restricted的调用）看起来太复杂了，不知在ORM查询中能不能做到？
实现菜单项排序的最佳实践是什么？
谢谢各位！
class MenuItem(db.Model):
"""
Menu item model
"""
__tablename__ = 'sa_menu_item'
id = db.Column(db.Integer, primary_key=True)
parent_id = db.Column(db.Integer, db.ForeignKey('sa_menu_item.id'))
text = db.Column(db.String(100))
view = db.Column(db.String(100))
icon = db.Column(db.String(50))
active = db.Column(db.Boolean)
children = db.relationship('MenuItem')
allowed_roles = db.relationship('Role',
secondary=menu_role
,backref='menu_item')
def __init__(self, text, view=None, icon=None):
self.text = text
self.view = view
self.icon = icon
def accessible_for(self, provided_set):
for role in self.allowed_roles:
if role.match(provided_set): return True
return False
def mark_restricted(self, lst, priv, not_allowed):
""" Adds menu items restricted by priv to not_allowed list """
if not self.accessible_for(priv):
if self in lst:
not_allowed.append(lst.index(self))
return
if self.children is not None:
for child in self.children:
if not child.accessible_for(priv):
self.children.remove(child)
else:
child.mark_restricted(lst, priv,not_allowed)
@classmethod
def generate_menu(cls, provided_set):
"""Generates menu based on provided_set of permissions"""
not_allowed = []
lst = MenuItem.query.filter(MenuItem.parent_id == None, MenuItem.active == True, MenuItem.children != None).all()
for item in lst:
item.mark_restricted(lst,provided_set,not_allowed)
lst = [i for j, i in enumerate(lst) if j not in not_allowed]
return lst