LightSwitch 2011 数据字段唯一性验证方案

LightSwitch 2011 数据字段唯一性验证方案

 


 

验证单表数据的某个字段不能输入重复值

 

设置实体字段唯一索引

如果不写代码,那么验证只会在用户提交[保存]数据后,会提示错误,很明显这样的用户体验并不好,因此还需要做以下步骤

添加自定义验证

 View Code

 
partial void UserName_Validate(EntityValidationResultsBuilder results)
        {
            // results.AddPropertyError("<错误消息>");
            bool duplicateExists = false;
           
            switch (this.Details.EntityState)
            {
                case EntityState.Added:
                    {
                        //基于页面未提交数据的验证
                        duplicateExists = (from item in DataWorkspace.ApplicationData.Details.GetChanges().AddedEntities.OfType<Employee>()
                                           where item.UserName == this.UserName && !string.IsNullOrEmpty(this.UserName)
                                           select item).Count() > 1 ? true : false;
                        //基于数据库的验证
                        if (!duplicateExists)
                        duplicateExists = (from Employee emp in DataWorkspace.ApplicationData.Employees.Cast<Employee>()
                                           where this.UserName != null &&
                                           string.Compare(emp.UserName, this.UserName.Trim(), StringComparison.InvariantCultureIgnoreCase) == 0
                                           select emp).Any();
                        break;
                    }

                case EntityState.Modified:
                    {
                        duplicateExists = (from item in DataWorkspace.ApplicationData.Details.GetChanges().ModifiedEntities.OfType<Employee>()
                                           where item.UserName == this.UserName && !string.IsNullOrEmpty(this.UserName)
                                           select item).Count() > 1 ? true : false;
                        if (!duplicateExists)
                        duplicateExists = (from Employee emp in DataWorkspace.ApplicationData.Employees.Cast<Employee>()
                                           where this.UserName != null &&
                                           string.Compare(emp.UserName, this.UserName.Trim(), StringComparison.InvariantCultureIgnoreCase) == 0
                                           select emp).Any();
                        break;
                    }
            }

            if (duplicateExists)
            {
                results.AddPropertyError(string.Format("该用户[{0}]已经存在。", UserName));

            }

运行结果如下

posted @ 2011-10-19 09:41  阿新  阅读(1697)  评论(1编辑  收藏  举报