当表单验证失败的查询字符串参数是错过了

本文介绍了当表单验证失败的查询字符串参数是错过了的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有以下URL的形式：CreateEntity？办公codeID = 5

当我发送的形式来验证，如果验证失败是它返回只是CreateEntity网址。没有办公室codeID = 5。

如果用户点击URL或F5输入 - 我的地盘失败 - 它需要缺少办公codeID参数。我可以将它保存到会话或其他存储。但我想有它的网址

我的观点：

[HTTPGET]        公共虚拟的ActionResult CreateEntity（INT？办公室codeID）        {            VAR模型=新CreateViewModel（）;            FillViewModel（型号，办公室codeID）;            返回视图（创建，模型）;        }[HttpPost]受保护的虚拟的ActionResult CreateEntity（TEditViewModel模型）        {            如果（ModelState.IsValid）            {              //做一些模型的东西，如果            }            返回视图（创建，模型）;        }

编辑。我的观点：

使用（Html.BeginForm（CreateEntity，雇员，FormMethod.Post，新{ENCTYPE =的multipart / form-data的}））{@ Html.HiddenFor（X =＆GT; x.Office codeID）＆LT; D​​IV＆GT;                @ Html.LabelFor（型号=＆GT; model.FirstName，CommonRes.FirstNameCol）                @ Html.TextBoxFor（型号=＆GT; model.FirstName，Model.FirstName）                @ Html.ValidationMessageFor（型号=＆GT; model.FirstName）            ＆LT; / DIV＆GT;            ＆LT; D​​IV＆GT;                @ Html.LabelFor（型号=＆GT; model.LastName，CommonRes.LastNameCol）                @ Html.TextBoxFor（型号=＆GT; model.LastName，Model.LastName）                @ Html.ValidationMessageFor（型号=＆GT; model.LastName）            ＆LT; / DIV＆GT;＆LT; D​​IV＆GT; ＆LT; D​​IV CLASS =输入文件名区＆GT;＆LT; / DIV＆GT;                    ＆LT;输入ID =协议类型=文件名称=协议/＆GT;＆LT; / DIV＆GT;

}

编辑2。添加：

@using（Html.BeginForm（CreateEntity，雇员，FormMethod.Post，新的办公{codeID = Model.Office codeID，ENCTYPE =的multipart / form-data的}））

Haven`t帮助。它产生以下形式：

＆LT;形式的行动=/ PhoneEmployee / CreateEntityENCTYPE =的multipart / form-data的方法=邮报的办公室codeID =5＆GT;

的解决方案是

＆LT;形式的行动=@ Url.Action（CreateEntity，员工）？office$c$cid=@Model.Office$c$cId~~VENCTYPE =的multipart / form-data的方法=后＆GT;

解决方案

问题是你的 HttpPost 动作没有的任何观念 ID 参数。如果你想支持一个类似的URL，然后做出行动签名支持该参数例如

[HTTPGET]公众的ActionResult CreateEntity（INT？办公室codeID）[HttpPost]公众的ActionResult CreateEntity（INT办公室codeID，EditViewModel模型）;

I have an form with following url: CreateEntity?officeCodeId=5

When I send form to validate and if validation is fail it returns just CreateEntity url. No officeCodeId=5.

if user click enter on URL or F5 - my site fail - it require missing officecodeId param. I can save it to the session or in the other storage. But I want to have it in the URL

My view:

[HttpGet] public virtual ActionResult CreateEntity(int? officeCodeId) { var model = new CreateViewModel(); FillViewModel(model, officeCodeId); return View("Create", model); } [HttpPost] protected virtual ActionResult CreateEntity(TEditViewModel model) { if (ModelState.IsValid) { //Do some model stuff if } return View("Create", model); }

EDIT. My View:

using (Html.BeginForm("CreateEntity", "Employee", FormMethod.Post, new { enctype = "multipart/form-data" })) { @Html.HiddenFor(x => x.OfficeCodeId) <div> @Html.LabelFor(model => model.FirstName, CommonRes.FirstNameCol) @Html.TextBoxFor(model => model.FirstName, Model.FirstName) @Html.ValidationMessageFor(model => model.FirstName) </div> <div> @Html.LabelFor(model => model.LastName, CommonRes.LastNameCol) @Html.TextBoxFor(model => model.LastName, Model.LastName) @Html.ValidationMessageFor(model => model.LastName) </div> <div> <div class="input-file-area"></div> <input id="Agreements" type="file" name="Agreements"/> </div>

}

Edit 2. Adding:

@using (Html.BeginForm("CreateEntity", "Employee", FormMethod.Post, new { officeCodeId = Model.OfficeCodeId, enctype = "multipart/form-data" }))

Haven`t help. It produce the following form:

<form action="/PhoneEmployee/CreateEntity" enctype="multipart/form-data" method="post" officecodeid="5">

Solution Is

<form action="@Url.Action("CreateEntity", "Employee")?officecodeid=@Model.OfficeCodeId" enctype="multipart/form-data" method="post">

解决方案

The problem is your HttpPost action doesn't have any notion of an id parameter. If you want to support a similar URL then make the action signature support that parameter e.g.

[HttpGet] public ActionResult CreateEntity(int? officeCodeId) [HttpPost] public ActionResult CreateEntity(int officeCodeId, EditViewModel model);