LightSwitch 2011 數據字段唯一性驗證方案

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驗證單表數據的某個字段不能輸入重復值

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設置實體字段唯一索引

如果不寫代碼，那么驗證只會在用戶提交[保存]數據后，會提示錯誤，很明顯這樣的用戶體驗并不好，因此還需要做以下步驟

添加自定義驗證

?View Code

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partial?void?UserName_Validate(EntityValidationResultsBuilder?results)
????????{
????????????//?results.AddPropertyError("<錯誤消息>");
????????????bool?duplicateExists?=?false;
???????????
????????????switch?(this.Details.EntityState)
????????????{
????????????????case?EntityState.Added:
????????????????????{
????????????????????????//基于頁面未提交數據的驗證
????????????????????????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;
????????????????????????//基于數據庫的驗證
????????????????????????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("該用戶[{0}]已經存在。",?UserName));

????????????}

運行結果如下

轉載于:https://www.cnblogs.com/neozhu/archive/2011/10/19/2217221.html

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