feat(role): 新增平台角色管理功能增强
- 权限表增加 available_for_role_types 字段,支持标记权限可用角色类型 - 权限列表和权限树接口支持按 available_for_role_type 过滤 - 新增角色状态切换接口 PUT /api/admin/roles/:id/status - 角色分配权限时验证权限的可用角色类型 - 完善数据库迁移脚本和单元测试 - 补充数据库迁移相关开发规范文档
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@@ -202,6 +202,9 @@ func (s *Service) List(ctx context.Context, req *model.PermissionListRequest) ([
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if req.Platform != "" {
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filters["platform"] = req.Platform
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}
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if req.AvailableForRoleType != nil {
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filters["available_for_role_type"] = *req.AvailableForRoleType
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}
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if req.ParentID != nil {
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filters["parent_id"] = *req.ParentID
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}
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@@ -213,35 +216,32 @@ func (s *Service) List(ctx context.Context, req *model.PermissionListRequest) ([
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}
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// GetTree 获取权限树
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func (s *Service) GetTree(ctx context.Context) ([]*model.PermissionTreeNode, error) {
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// 获取所有权限
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permissions, err := s.permissionStore.GetAll(ctx)
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func (s *Service) GetTree(ctx context.Context, availableForRoleType *int) ([]*model.PermissionTreeNode, error) {
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permissions, err := s.permissionStore.GetAll(ctx, availableForRoleType)
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if err != nil {
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return nil, fmt.Errorf("获取权限列表失败: %w", err)
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}
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// 构建树结构
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return buildPermissionTree(permissions), nil
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}
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// buildPermissionTree 构建权限树
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func buildPermissionTree(permissions []*model.Permission) []*model.PermissionTreeNode {
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// 转换为节点映射
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nodeMap := make(map[uint]*model.PermissionTreeNode)
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for _, p := range permissions {
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nodeMap[p.ID] = &model.PermissionTreeNode{
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ID: p.ID,
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PermName: p.PermName,
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PermCode: p.PermCode,
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PermType: p.PermType,
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Platform: p.Platform,
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URL: p.URL,
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Sort: p.Sort,
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Children: make([]*model.PermissionTreeNode, 0),
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ID: p.ID,
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PermName: p.PermName,
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PermCode: p.PermCode,
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PermType: p.PermType,
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Platform: p.Platform,
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AvailableForRoleTypes: p.AvailableForRoleTypes,
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URL: p.URL,
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Sort: p.Sort,
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Children: make([]*model.PermissionTreeNode, 0),
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}
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}
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// 构建树
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var roots []*model.PermissionTreeNode
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for _, p := range permissions {
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node := nodeMap[p.ID]
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@@ -250,7 +250,6 @@ func buildPermissionTree(permissions []*model.Permission) []*model.PermissionTre
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} else if parent, ok := nodeMap[*p.ParentID]; ok {
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parent.Children = append(parent.Children, node)
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} else {
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// 如果找不到父节点,作为根节点处理
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roots = append(roots, node)
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}
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}
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