204 lines
7.4 KiB
Python
204 lines
7.4 KiB
Python
from app.models.system import SysMenu, SysElement, SysRolePermission
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from app.extensions import db
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from sqlalchemy.exc import SQLAlchemyError
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class PermissionService:
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@staticmethod
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def get_permission_tree():
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"""
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获取完整的权限树（菜单嵌套菜单 + 菜单包含元素）
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供前端权限配置页面展示
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"""
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# 1. 获取所有菜单 (按 parent_id 和 sort_order 排序，保证父子处理顺序)
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menus = SysMenu.query.order_by(SysMenu.parent_id, SysMenu.sort_order).all()
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# 2. 获取所有元素
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elements = SysElement.query.all()
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# --- 核心逻辑：构建树形结构 ---
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# 3. 创建一个 lookup 字典，方便通过 ID 查找菜单节点
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# 同时将 SQLAlchemy 对象转为字典，方便后续操作
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menu_map = {}
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for m in menus:
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m_dict = m.to_dict()
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m_dict['children'] = [] # 初始化 children
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menu_map[m.id] = m_dict
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# 4. 创建 code 到 id 的映射，用于把 element 挂载到 menu 上
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# 因为 SysElement 关联的是 menu_code，而不是 menu_id
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code_to_id = {m.code: m.id for m in menus}
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# 5. 将元素 (Elements) 挂载到对应的菜单 (Menu) 下
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for el in elements:
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# 找到该元素所属菜单的 ID
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parent_menu_id = code_to_id.get(el.menu_code)
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if parent_menu_id and parent_menu_id in menu_map:
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el_dict = el.to_dict()
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# 标记类型为 element，前端 transformData 需要用到
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el_dict['type'] = 'element'
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menu_map[parent_menu_id]['children'].append(el_dict)
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# 6. 将子菜单挂载到父菜单下，并构建最终的树
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tree_data = []
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for m in menus:
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current_node = menu_map[m.id]
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if m.parent_id == 0 or m.parent_id is None:
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# 如果是顶级菜单，直接放入结果集
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tree_data.append(current_node)
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else:
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# 如果是子菜单，找到它的父级，把它塞进父级的 children 里
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if m.parent_id in menu_map:
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menu_map[m.parent_id]['children'].append(current_node)
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else:
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# 如果找不到父级（比如父级被删了），为了防止数据丢失，暂时作为顶级显示
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tree_data.append(current_node)
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return tree_data
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@staticmethod
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def get_role_permissions(role_code):
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"""获取指定角色拥有的所有权限Code"""
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try:
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# === 新增逻辑：超级管理员上帝模式 ===
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if role_code == 'SUPER_ADMIN':
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# 直接获取所有菜单和元素，无视配置表
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all_menus = [m.code for m in SysMenu.query.all()]
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all_elements = [e.code for e in SysElement.query.all()]
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return {
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'menus': all_menus,
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'elements': all_elements
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}
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# =================================
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perms = SysRolePermission.query.filter_by(role_code=role_code).all()
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menu_codes = []
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element_codes = []
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for p in perms:
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# 这里假设你的数据库存的是 target_code
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if p.type == 'menu':
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menu_codes.append(p.target_code)
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else:
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element_codes.append(p.target_code)
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# 前端 handleRoleSelect 会合并这两个数组，所以分开返回没问题
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return {
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'menus': menu_codes,
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'elements': element_codes
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}
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except Exception as e:
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# 记录日志或处理错误
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print(f"Error fetching role permissions: {e}")
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return {'menus': [], 'elements': []}
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@staticmethod
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def assign_permissions(role_code, permissions):
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"""
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保存角色的权限
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permissions: 前端传来的 list，混合了 menu_code 和 element_code
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"""
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if not role_code:
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raise ValueError("角色代码不能为空")
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session = db.session
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try:
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# 1. 开启事务 (Flask-SQLAlchemy 自动管理，但明确逻辑更好)
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# 2. 删除该角色旧的所有权限
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SysRolePermission.query.filter_by(role_code=role_code).delete()
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# 3. 准备新数据
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if permissions:
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# 3.1 去重
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unique_codes = set(permissions)
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# 3.2 预加载所有 Menu Code，用于区分是 Menu 还是 Element
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# 这一步很重要，因为 SysRolePermission 表需要 type 字段
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all_menu_codes = {res[0] for res in session.query(SysMenu.code).all()}
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new_records = []
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for code in unique_codes:
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if not code: continue
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# 判断类型：如果 code 存在于菜单表中，就是 menu，否则就是 element
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p_type = 'menu' if code in all_menu_codes else 'element'
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new_records.append(SysRolePermission(
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role_code=role_code,
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target_code=code,
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type=p_type
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))
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# 3.3 批量插入
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if new_records:
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session.add_all(new_records)
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# 4. 提交
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session.commit()
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return True
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except SQLAlchemyError as e:
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session.rollback()
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raise e
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except Exception as e:
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session.rollback()
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raise e
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@staticmethod
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def init_audit_menu():
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"""
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初始化审计日志菜单和超级管理员权限
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防重复：只插入不存在的记录
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"""
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try:
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# 1. 检查并创建审计日志菜单
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menu_code = 'system_audit'
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existing_menu = SysMenu.query.filter_by(code=menu_code).first()
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if not existing_menu:
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new_menu = SysMenu(
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parent_id=0,
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name='审计日志',
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code=menu_code,
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path='/system/audit',
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sort_order=110,
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is_visible=True
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)
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db.session.add(new_menu)
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db.session.flush() # 获取新插入的 ID
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print(f"✅ 审计日志菜单已创建 (code: {menu_code})")
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else:
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print(f"ℹ️ 审计日志菜单已存在 (code: {menu_code})")
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# 2. 为超级管理员赋予审计日志菜单权限
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role_code = 'SUPER_ADMIN'
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existing_perm = SysRolePermission.query.filter_by(
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role_code=role_code,
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target_code=menu_code
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).first()
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if not existing_perm:
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new_perm = SysRolePermission(
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role_code=role_code,
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target_code=menu_code,
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type='menu'
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)
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db.session.add(new_perm)
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print(f"✅ 超级管理员已赋予审计日志权限")
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else:
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print(f"ℹ️ 超级管理员已拥有审计日志权限")
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# 3. 提交
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db.session.commit()
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return True
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except Exception as e:
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db.session.rollback()
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print(f"❌ 初始化审计日志菜单失败: {str(e)}")
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raise e
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