532 lines
22 KiB
Python
532 lines
22 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. 获取所有可见菜单 (按 parent_id 和 sort_order 排序，保证父子处理顺序)
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menus = SysMenu.query.filter(SysMenu.is_visible == True).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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# 直接获取所有可见菜单和元素，无视配置表
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all_menus = [m.code for m in SysMenu.query.filter(SysMenu.is_visible == True).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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@staticmethod
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def init_stocktake_menus():
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"""
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初始化盘点管理菜单和权限
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包括：顶级菜单、盲盘作业、盈亏调整、以及操作权限元素
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"""
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try:
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role_code = 'SUPER_ADMIN'
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# 1. 创建顶级菜单：盘点管理
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stocktake_mgmt_code = 'stocktake_mgmt'
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stocktake_menu = SysMenu.query.filter_by(code=stocktake_mgmt_code).first()
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if not stocktake_menu:
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stocktake_menu = SysMenu(
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parent_id=0,
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name='盘点管理',
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code=stocktake_mgmt_code,
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path='/stocktake',
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sort_order=30,
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is_visible=True
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)
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db.session.add(stocktake_menu)
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db.session.flush()
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print(f"✅ 盘点管理顶级菜单已创建")
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else:
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print(f"ℹ️ 盘点管理顶级菜单已存在")
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# 2. 创建子菜单：盲盘作业
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stocktake_op_code = 'inventory_stocktake'
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stocktake_op_menu = SysMenu.query.filter_by(code=stocktake_op_code).first()
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if not stocktake_op_menu:
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stocktake_op_menu = SysMenu(
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parent_id=stocktake_menu.id,
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name='盲盘作业',
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code=stocktake_op_code,
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path='/stocktake/operation',
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sort_order=1,
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is_visible=True
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)
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db.session.add(stocktake_op_menu)
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db.session.flush()
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print(f"✅ 盲盘作业菜单已创建")
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else:
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print(f"ℹ️ 盲盘作业菜单已存在")
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# 3. 为盲盘作业添加操作权限元素
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stocktake_op_element = SysElement.query.filter_by(
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menu_code=stocktake_op_code,
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code='inventory_stocktake:operation'
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).first()
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if not stocktake_op_element:
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stocktake_op_element = SysElement(
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menu_code=stocktake_op_code,
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name='盲盘操作',
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code='inventory_stocktake:operation',
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element_type='operation'
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)
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db.session.add(stocktake_op_element)
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print(f"✅ 盲盘作业操作权限已创建")
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else:
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print(f"ℹ️ 盲盘作业操作权限已存在")
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# 4. 创建子菜单：盈亏调整
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adjustment_code = 'stock_adjustment'
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adjustment_menu = SysMenu.query.filter_by(code=adjustment_code).first()
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if not adjustment_menu:
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adjustment_menu = SysMenu(
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parent_id=stocktake_menu.id,
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name='盈亏调整',
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code=adjustment_code,
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path='/stocktake/adjustment',
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sort_order=2,
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is_visible=True
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)
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db.session.add(adjustment_menu)
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db.session.flush()
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print(f"✅ 盈亏调整菜单已创建")
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else:
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print(f"ℹ️ 盈亏调整菜单已存在")
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# 5. 为盈亏调整添加列表权限元素 (stock_adjustment:list)
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adjustment_list_element = SysElement.query.filter_by(
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menu_code=adjustment_code,
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code='stock_adjustment:list'
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).first()
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if not adjustment_list_element:
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adjustment_list_element = SysElement(
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menu_code=adjustment_code,
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name='盈亏列表',
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code='stock_adjustment:list',
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element_type='element'
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)
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db.session.add(adjustment_list_element)
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print(f"✅ 盈亏调整列表权限已创建")
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else:
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print(f"ℹ️ 盈亏调整列表权限已存在")
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# 6. 为盈亏调整添加操作权限元素 (stock_adjustment:operation)
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adjustment_op_element = SysElement.query.filter_by(
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menu_code=adjustment_code,
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code='stock_adjustment:operation'
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).first()
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if not adjustment_op_element:
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adjustment_op_element = SysElement(
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menu_code=adjustment_code,
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name='盈亏操作',
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code='stock_adjustment:operation',
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element_type='operation'
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)
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db.session.add(adjustment_op_element)
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print(f"✅ 盈亏调整操作权限已创建")
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else:
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print(f"ℹ️ 盈亏调整操作权限已存在")
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# 7. 为超级管理员分配所有盘点相关权限
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menu_codes = [stocktake_mgmt_code, stocktake_op_code, adjustment_code]
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for mc in menu_codes:
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existing_perm = SysRolePermission.query.filter_by(
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role_code=role_code,
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target_code=mc,
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type='menu'
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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=mc,
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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 {mc} 菜单权限")
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# 8. 分配操作权限
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op_codes = ['inventory_stocktake:operation', 'stock_adjustment:list', 'stock_adjustment:operation']
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for oc in op_codes:
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existing_perm = SysRolePermission.query.filter_by(
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role_code=role_code,
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target_code=oc,
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type='element'
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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=oc,
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type='element'
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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 {oc} 操作权限")
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# 9. 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"❌ 初始化盘点管理菜单失败: {str(e)}")
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raise e
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@staticmethod
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def cleanup_legacy_stocktake_menus():
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"""
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清理残留的旧版库存盘点菜单
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如果数据库中存在挂在入库管理下的旧版库存盘点菜单，清理掉
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"""
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try:
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# 查找可能存在的旧版库存盘点菜单（入库管理下的 stocktake）
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# 旧版可能在入库管理 (path like '%inventory%stocktake') 或者 code 包含 stocktake 但不是新的
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legacy_menus = SysMenu.query.filter(
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SysMenu.code.in_(['stocktake', 'inventory_stocktake_old'])
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).all()
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for menu in legacy_menus:
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# 删除关联的权限
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SysRolePermission.query.filter_by(target_code=menu.code).delete()
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# 删除关联的元素
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SysElement.query.filter_by(menu_code=menu.code).delete()
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# 删除菜单
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db.session.delete(menu)
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print(f"🗑️ 已清理旧版库存盘点菜单: {menu.code} ({menu.name})")
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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"⚠️ 清理旧版菜单失败: {str(e)}")
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return False
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@staticmethod
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def init_all_menus():
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"""
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初始化所有菜单的层级结构，确保权限配置页面显示正确的树形结构
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按照侧边栏顺序：基础信息 -> 入库 -> 盘点 -> 出库 -> BOM -> 借库 -> 报废 -> 系统
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"""
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try:
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role_code = 'SUPER_ADMIN'
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# 定义菜单结构 (code, name, path, parent_code, sort_order)
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menu_defs = [
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# 顶级菜单 (按侧边栏顺序)
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('material_mgmt', '基础信息管理', '/material', None, 10),
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('inventory_mgmt', '入库管理', '/inventory', None, 20),
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('stocktake_mgmt', '盘点管理', '/stocktake', None, 30),
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('outbound_mgmt', '出库管理', '/outbound', None, 40),
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('bom_mgmt', 'BOM管理', '/bom', None, 50),
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('operation_mgmt', '借库管理', '/operation', None, 60),
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('scrap_mgmt', '报废管理', '/scrap', None, 70),
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('system_mgmt', '系统管理', '/system', None, 80),
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# 基础信息子菜单
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('material_base', '基础信息', '/material/index', 'material_mgmt', 1),
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# 入库管理子菜单
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('inbound_buy', '采购入库', '/inventory/buy', 'inventory_mgmt', 1),
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('inbound_semi', '半成品入库', '/inventory/semi', 'inventory_mgmt', 2),
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('inbound_product', '成品入库', '/inventory/product', 'inventory_mgmt', 3),
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('inbound_service', '服务权益', '/inventory/service', 'inventory_mgmt', 4),
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('inbound_summary', '入库记录', '/inventory/summary', 'inventory_mgmt', 5),
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# 盘点管理子菜单
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('inventory_stocktake', '盲盘作业', '/stocktake/operation', 'stocktake_mgmt', 1),
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('stock_adjustment', '盈亏调整', '/stocktake/adjustment', 'stocktake_mgmt', 2),
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# 出库管理子菜单
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('outbound_selection', '出库选单', '/outbound/selection', 'outbound_mgmt', 1),
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('outbound_create', '扫码出库', '/outbound/create', 'outbound_mgmt', 2),
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('outbound_list', '出库记录', '/outbound/index', 'outbound_mgmt', 3),
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# BOM管理子菜单
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('bom_manage', 'BOM配方管理', '/bom/manage', 'bom_mgmt', 1),
|
||
|
||
# 借库管理子菜单
|
||
('op_borrow', '借库操作', '/operation/borrow', 'operation_mgmt', 1),
|
||
('op_return', '归还操作', '/operation/repair', 'operation_mgmt', 2),
|
||
('op_records', '借还记录', '/operation/records', 'operation_mgmt', 3),
|
||
|
||
# 报废管理子菜单
|
||
('scrap_create', '新建报废', '/scrap/create', 'scrap_mgmt', 1),
|
||
('scrap_list', '报废记录', '/scrap/index', 'scrap_mgmt', 2),
|
||
|
||
# 系统管理子菜单
|
||
('system_user', '员工账号管理', '/system/user-create', 'system_mgmt', 1),
|
||
('system_permission', '权限分配', '/system/permission', 'system_mgmt', 2),
|
||
('system_audit', '审计日志', '/system/audit', 'system_mgmt', 3),
|
||
]
|
||
|
||
# 第一步：清理根级别的冗余子菜单（这些本应是子节点，但可能之前被错误地创建为根节点）
|
||
child_codes = [m[0] for m in menu_defs if m[3] is not None] # 所有子菜单的code
|
||
orphaned_menus = SysMenu.query.filter(
|
||
SysMenu.code.in_(child_codes),
|
||
(SysMenu.parent_id == 0) | (SysMenu.parent_id.is_(None))
|
||
).all()
|
||
for menu in orphaned_menus:
|
||
print(f"🗑️ 清理根级别冗余菜单: {menu.code} ({menu.name})")
|
||
# 删除关联的权限
|
||
SysRolePermission.query.filter_by(target_code=menu.code).delete()
|
||
db.session.delete(menu)
|
||
|
||
# 第二步：清理重复菜单（同一个 code 存在多条记录，保留 ID 最小的）
|
||
# 先找出所有可能重复的 code
|
||
duplicate_check = db.session.query(
|
||
SysMenu.code,
|
||
func.count(SysMenu.id).label('cnt')
|
||
).group_by(SysMenu.code).having(func.count(SysMenu.id) > 1).all()
|
||
|
||
for code, cnt in duplicate_check:
|
||
# 获取该 code 的所有记录，按 id 排序，保留第一条，删除其余
|
||
duplicates = SysMenu.query.filter_by(code=code).order_by(SysMenu.id).all()
|
||
# 保留第一条，删除其他
|
||
for dup in duplicates[1:]:
|
||
print(f"🗑️ 清理重复菜单: {dup.code} (id={dup.id}, name={dup.name})")
|
||
SysRolePermission.query.filter_by(target_code=dup.code).delete()
|
||
SysElement.query.filter_by(menu_code=dup.code).delete()
|
||
db.session.delete(dup)
|
||
|
||
# 第三步：强制重新设置所有子菜单的 parent_id，确保没有遗漏
|
||
# 先将所有子菜单的 parent_id 设为 None，然后重新设置
|
||
SysMenu.query.filter(SysMenu.code.in_(child_codes)).update({SysMenu.parent_id: None})
|
||
|
||
# 创建或更新菜单
|
||
menu_map = {} # code -> menu obj
|
||
|
||
for code, name, path, parent_code, sort_order in menu_defs:
|
||
menu = SysMenu.query.filter_by(code=code).first()
|
||
if not menu:
|
||
menu = SysMenu(code=code, name=name, path=path, sort_order=sort_order, is_visible=True)
|
||
db.session.add(menu)
|
||
db.session.flush()
|
||
print(f"✅ 菜单已创建: {name} ({code})")
|
||
else:
|
||
# 更新已有菜单的属性
|
||
menu.name = name
|
||
menu.path = path
|
||
menu.sort_order = sort_order
|
||
|
||
menu_map[code] = menu
|
||
|
||
# 设置 parent_id
|
||
for code, name, path, parent_code, sort_order in menu_defs:
|
||
if parent_code and parent_code in menu_map:
|
||
menu = menu_map[code]
|
||
parent = menu_map[parent_code]
|
||
menu.parent_id = parent.id
|
||
|
||
# 为超级管理员分配所有菜单权限
|
||
for code, name, path, parent_code, sort_order in menu_defs:
|
||
if parent_code is None: # 只分配顶级菜单
|
||
existing_perm = SysRolePermission.query.filter_by(
|
||
role_code=role_code,
|
||
target_code=code,
|
||
type='menu'
|
||
).first()
|
||
if not existing_perm:
|
||
new_perm = SysRolePermission(
|
||
role_code=role_code,
|
||
target_code=code,
|
||
type='menu'
|
||
)
|
||
db.session.add(new_perm)
|
||
|
||
db.session.commit()
|
||
print(f"✅ 所有菜单初始化完成")
|
||
return True
|
||
|
||
except Exception as e:
|
||
db.session.rollback()
|
||
print(f"❌ 初始化菜单失败: {str(e)}")
|
||
raise e
|