from app.models.system import SysMenu, SysElement, SysRolePermission from app.extensions import db from sqlalchemy.exc import SQLAlchemyError class PermissionService: @staticmethod def get_permission_tree(): """ 获取完整的权限树（菜单嵌套菜单 + 菜单包含元素） 供前端权限配置页面展示 """ # 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) menus = SysMenu.query.filter(SysMenu.is_visible == True).order_by(SysMenu.parent_id, SysMenu.sort_order).all() # 2. 获取所有元素 elements = SysElement.query.all() # --- 核心逻辑：构建树形结构 --- # 3. 创建一个 lookup 字典，方便通过 ID 查找菜单节点 # 同时将 SQLAlchemy 对象转为字典，方便后续操作 menu_map = {} for m in menus: m_dict = m.to_dict() m_dict['children'] = [] # 初始化 children menu_map[m.id] = m_dict # 4. 创建 code 到 id 的映射，用于把 element 挂载到 menu 上 # 因为 SysElement 关联的是 menu_code，而不是 menu_id code_to_id = {m.code: m.id for m in menus} # 5. 将元素 (Elements) 挂载到对应的菜单 (Menu) 下 for el in elements: # 找到该元素所属菜单的 ID parent_menu_id = code_to_id.get(el.menu_code) if parent_menu_id and parent_menu_id in menu_map: el_dict = el.to_dict() # 标记类型为 element，前端 transformData 需要用到 el_dict['type'] = 'element' menu_map[parent_menu_id]['children'].append(el_dict) # 6. 将子菜单挂载到父菜单下，并构建最终的树 tree_data = [] for m in menus: current_node = menu_map[m.id] if m.parent_id == 0 or m.parent_id is None: # 如果是顶级菜单，直接放入结果集 tree_data.append(current_node) else: # 如果是子菜单，找到它的父级，把它塞进父级的 children 里 if m.parent_id in menu_map: menu_map[m.parent_id]['children'].append(current_node) else: # 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 tree_data.append(current_node) return tree_data @staticmethod def get_role_permissions(role_code): """获取指定角色拥有的所有权限Code""" try: # === 新增逻辑：超级管理员上帝模式 === if role_code == 'SUPER_ADMIN': # 直接获取所有可见菜单和元素，无视配置表 all_menus = [m.code for m in SysMenu.query.filter(SysMenu.is_visible == True).all()] all_elements = [e.code for e in SysElement.query.all()] return { 'menus': all_menus, 'elements': all_elements } # ================================= perms = SysRolePermission.query.filter_by(role_code=role_code).all() menu_codes = [] element_codes = [] for p in perms: # 这里假设你的数据库存的是 target_code if p.type == 'menu': menu_codes.append(p.target_code) else: element_codes.append(p.target_code) # 前端 handleRoleSelect 会合并这两个数组，所以分开返回没问题 return { 'menus': menu_codes, 'elements': element_codes } except Exception as e: # 记录日志或处理错误 print(f"Error fetching role permissions: {e}") return {'menus': [], 'elements': []} @staticmethod def assign_permissions(role_code, permissions): """ 保存角色的权限 permissions: 前端传来的 list，混合了 menu_code 和 element_code """ if not role_code: raise ValueError("角色代码不能为空") session = db.session try: # 1. 开启事务 (Flask-SQLAlchemy 自动管理，但明确逻辑更好) # 2. 删除该角色旧的所有权限 SysRolePermission.query.filter_by(role_code=role_code).delete() # 3. 准备新数据 if permissions: # 3.1 去重 unique_codes = set(permissions) # 3.2 预加载所有 Menu Code，用于区分是 Menu 还是 Element # 这一步很重要，因为 SysRolePermission 表需要 type 字段 all_menu_codes = {res[0] for res in session.query(SysMenu.code).all()} new_records = [] for code in unique_codes: if not code: continue # 判断类型：如果 code 存在于菜单表中，就是 menu，否则就是 element p_type = 'menu' if code in all_menu_codes else 'element' new_records.append(SysRolePermission( role_code=role_code, target_code=code, type=p_type )) # 3.3 批量插入 if new_records: session.add_all(new_records) # 4. 提交 session.commit() return True except SQLAlchemyError as e: session.rollback() raise e except Exception as e: session.rollback() raise e @staticmethod def init_audit_menu(): """ 初始化审计日志菜单和超级管理员权限 防重复：只插入不存在的记录 """ try: # 1. 检查并创建审计日志菜单 menu_code = 'system_audit' existing_menu = SysMenu.query.filter_by(code=menu_code).first() if not existing_menu: new_menu = SysMenu( parent_id=0, name='审计日志', code=menu_code, path='/system/audit', sort_order=110, is_visible=True ) db.session.add(new_menu) db.session.flush() # 获取新插入的 ID print(f"✅ 审计日志菜单已创建 (code: {menu_code})") else: print(f"ℹ️ 审计日志菜单已存在 (code: {menu_code})") # 2. 为超级管理员赋予审计日志菜单权限 role_code = 'SUPER_ADMIN' existing_perm = SysRolePermission.query.filter_by( role_code=role_code, target_code=menu_code ).first() if not existing_perm: new_perm = SysRolePermission( role_code=role_code, target_code=menu_code, type='menu' ) db.session.add(new_perm) print(f"✅ 超级管理员已赋予审计日志权限") else: print(f"ℹ️ 超级管理员已拥有审计日志权限") # 3. 提交 db.session.commit() return True except Exception as e: db.session.rollback() print(f"❌ 初始化审计日志菜单失败: {str(e)}") raise e @staticmethod def init_stocktake_menus(): """ 初始化盘点管理菜单和权限 包括：顶级菜单、盲盘作业、盈亏调整、以及操作权限元素 """ try: role_code = 'SUPER_ADMIN' # 1. 创建顶级菜单：盘点管理 stocktake_mgmt_code = 'stocktake_mgmt' stocktake_menu = SysMenu.query.filter_by(code=stocktake_mgmt_code).first() if not stocktake_menu: stocktake_menu = SysMenu( parent_id=0, name='盘点管理', code=stocktake_mgmt_code, path='/stocktake', sort_order=30, is_visible=True ) db.session.add(stocktake_menu) db.session.flush() print(f"✅ 盘点管理顶级菜单已创建") else: print(f"ℹ️ 盘点管理顶级菜单已存在") # 2. 创建子菜单：盲盘作业 stocktake_op_code = 'inventory_stocktake' stocktake_op_menu = SysMenu.query.filter_by(code=stocktake_op_code).first() if not stocktake_op_menu: stocktake_op_menu = SysMenu( parent_id=stocktake_menu.id, name='盲盘作业', code=stocktake_op_code, path='/stocktake/operation', sort_order=1, is_visible=True ) db.session.add(stocktake_op_menu) db.session.flush() print(f"✅ 盲盘作业菜单已创建") else: print(f"ℹ️ 盲盘作业菜单已存在") # 3. 为盲盘作业添加操作权限元素 stocktake_op_element = SysElement.query.filter_by( menu_code=stocktake_op_code, code='inventory_stocktake:operation' ).first() if not stocktake_op_element: stocktake_op_element = SysElement( menu_code=stocktake_op_code, name='盲盘操作', code='inventory_stocktake:operation', element_type='operation' ) db.session.add(stocktake_op_element) print(f"✅ 盲盘作业操作权限已创建") else: print(f"ℹ️ 盲盘作业操作权限已存在") # 4. 创建子菜单：盈亏调整 adjustment_code = 'stock_adjustment' adjustment_menu = SysMenu.query.filter_by(code=adjustment_code).first() if not adjustment_menu: adjustment_menu = SysMenu( parent_id=stocktake_menu.id, name='盈亏调整', code=adjustment_code, path='/stocktake/adjustment', sort_order=2, is_visible=True ) db.session.add(adjustment_menu) db.session.flush() print(f"✅ 盈亏调整菜单已创建") else: print(f"ℹ️ 盈亏调整菜单已存在") # 5. 为盈亏调整添加列表权限元素 (stock_adjustment:list) adjustment_list_element = SysElement.query.filter_by( menu_code=adjustment_code, code='stock_adjustment:list' ).first() if not adjustment_list_element: adjustment_list_element = SysElement( menu_code=adjustment_code, name='盈亏列表', code='stock_adjustment:list', element_type='element' ) db.session.add(adjustment_list_element) print(f"✅ 盈亏调整列表权限已创建") else: print(f"ℹ️ 盈亏调整列表权限已存在") # 6. 为盈亏调整添加操作权限元素 (stock_adjustment:operation) adjustment_op_element = SysElement.query.filter_by( menu_code=adjustment_code, code='stock_adjustment:operation' ).first() if not adjustment_op_element: adjustment_op_element = SysElement( menu_code=adjustment_code, name='盈亏操作', code='stock_adjustment:operation', element_type='operation' ) db.session.add(adjustment_op_element) print(f"✅ 盈亏调整操作权限已创建") else: print(f"ℹ️ 盈亏调整操作权限已存在") # 7. 为超级管理员分配所有盘点相关权限 menu_codes = [stocktake_mgmt_code, stocktake_op_code, adjustment_code] for mc in menu_codes: existing_perm = SysRolePermission.query.filter_by( role_code=role_code, target_code=mc, type='menu' ).first() if not existing_perm: new_perm = SysRolePermission( role_code=role_code, target_code=mc, type='menu' ) db.session.add(new_perm) print(f"✅ 超级管理员已赋予 {mc} 菜单权限") # 8. 分配操作权限 op_codes = ['inventory_stocktake:operation', 'stock_adjustment:list', 'stock_adjustment:operation'] for oc in op_codes: existing_perm = SysRolePermission.query.filter_by( role_code=role_code, target_code=oc, type='element' ).first() if not existing_perm: new_perm = SysRolePermission( role_code=role_code, target_code=oc, type='element' ) db.session.add(new_perm) print(f"✅ 超级管理员已赋予 {oc} 操作权限") # 9. 提交 db.session.commit() return True except Exception as e: db.session.rollback() print(f"❌ 初始化盘点管理菜单失败: {str(e)}") raise e @staticmethod def cleanup_legacy_stocktake_menus(): """ 清理残留的旧版库存盘点菜单 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 """ try: # 查找可能存在的旧版库存盘点菜单（入库管理下的 stocktake） # 旧版可能在入库管理 (path like '%inventory%stocktake') 或者 code 包含 stocktake 但不是新的 legacy_menus = SysMenu.query.filter( SysMenu.code.in_(['stocktake', 'inventory_stocktake_old']) ).all() for menu in legacy_menus: # 删除关联的权限 SysRolePermission.query.filter_by(target_code=menu.code).delete() # 删除关联的元素 SysElement.query.filter_by(menu_code=menu.code).delete() # 删除菜单 db.session.delete(menu) print(f"🗑️ 已清理旧版库存盘点菜单: {menu.code} ({menu.name})") db.session.commit() return True except Exception as e: db.session.rollback() print(f"⚠️ 清理旧版菜单失败: {str(e)}") return False @staticmethod def init_all_menus(): """ 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 按照侧边栏顺序：基础信息 -> 入库 -> 盘点 -> 出库 -> BOM -> 借库 -> 报废 -> 系统 """ try: role_code = 'SUPER_ADMIN' # 定义菜单结构 (code, name, path, parent_code, sort_order) menu_defs = [ # 顶级菜单 (按侧边栏顺序) ('material_mgmt', '基础信息管理', '/material', None, 10), ('inventory_mgmt', '入库管理', '/inventory', None, 20), ('stocktake_mgmt', '盘点管理', '/stocktake', None, 30), ('outbound_mgmt', '出库管理', '/outbound', None, 40), ('bom_mgmt', 'BOM管理', '/bom', None, 50), ('operation_mgmt', '借库管理', '/operation', None, 60), ('scrap_mgmt', '报废管理', '/scrap', None, 70), ('system_mgmt', '系统管理', '/system', None, 80), # 基础信息子菜单 ('material_base', '基础信息', '/material/index', 'material_mgmt', 1), # 入库管理子菜单 ('inbound_buy', '采购入库', '/inventory/buy', 'inventory_mgmt', 1), ('inbound_semi', '半成品入库', '/inventory/semi', 'inventory_mgmt', 2), ('inbound_product', '成品入库', '/inventory/product', 'inventory_mgmt', 3), ('inbound_service', '服务权益', '/inventory/service', 'inventory_mgmt', 4), ('inbound_summary', '入库记录', '/inventory/summary', 'inventory_mgmt', 5), # 盘点管理子菜单 ('inventory_stocktake', '盲盘作业', '/stocktake/operation', 'stocktake_mgmt', 1), ('stock_adjustment', '盈亏调整', '/stocktake/adjustment', 'stocktake_mgmt', 2), # 出库管理子菜单 ('outbound_selection', '出库选单', '/outbound/selection', 'outbound_mgmt', 1), ('outbound_create', '扫码出库', '/outbound/create', 'outbound_mgmt', 2), ('outbound_list', '出库记录', '/outbound/index', 'outbound_mgmt', 3), # BOM管理子菜单 ('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