当针对模型类验证模型时，我试图将验证消息显示回部分 View 。但是当失败时，我不知道如何将验证错误消息从 POST 的位置显示回 VIEW被解雇 这是部分 View 的代码 (_CreateCustomer)
@model Banking.Models.Customer
@Html.ValidationSummary(true)
<fieldset>
@using (Html.BeginForm("CreateCustomer", "Customer",FormMethod.Post))
{
@Html.ValidationSummary(true)
<div id="cusCreate">
<table id="createCustomer">
<tr><td>Please enter ther details of the Customer </td></tr>
<tr>
<td>First Name : </td>
<td>
@Html.TextBoxFor(model => model.FirstName)
@Html.ValidationMessageFor(model => model.FirstName)
</td>
<td>Last Name : </td>
<td>
@Html.TextBoxFor(model => model.LastName)
@Html.ValidationMessageFor(model => model.LastName)
</td>
</tr>
<tr>
<td>Address : </td>
<td>
@Html.TextAreaFor(model => model.Address)
@Html.ValidationMessageFor(model => model.Address)
</td>
<td>City : </td>
<td>
@Html.TextBoxFor(model => model.City)
@Html.ValidationMessageFor(model => model.City)
</td>
</tr>
<tr>
<td>State : </td>
<td>
@Html.TextBoxFor(model => model.State)
@Html.ValidationMessageFor(model => model.State)
</td>
<td>Pincode :</td>
<td>
@Html.EditorFor(model => model.Pincode)
@Html.ValidationMessageFor(model => model.Pincode)
</td>
</tr>
<tr>
<td>Phone : </td>
<td>
@Html.EditorFor(model => model.Phone)
@Html.ValidationMessageFor(model => model.Phone)
</td>
<td>Email :</td>
<td>
@Html.TextBoxFor(model => model.Email)
@Html.ValidationMessageFor(model => model.Email)
</td>
</tr>
<tr>
<td style="align-content:center" colspan="4"> </td>
</tr>
</table>
<input type="submit" value="CreateCustomer" id="cusSubBtn" />
</div>
}
</fieldset>
以下是加载局部 View 的 View 代码
@{
ViewBag.Title = "CustomerMain";
Layout = "~/Views/Shared/_Layout.cshtml";
}
@Scripts.Render("~/bundles/jquery")
<html>
<head>
<script type="text/javascript">
$(document).ready(function () {
$('#CusCreatebtn').click(function (event) {
$("#CusDynamicPortion").hide();
$("#CusDynamicPortion").load("/Customer/CreateCustomerData");
$("#CusDynamicPortion").show();
});
});
</script>
</head>
<body >
<div id="CusDynamicPortion">
<table id="customeroptions" style="width:100%">
<tr>
<td><input type="button" id="CusCreatebtn" value="Add New Customer" /> </td>
<td><input type="button" id="CusViewbtn" value="View Customer" /> </td>
</tr>
</table>
</div>
</body>
</html>
对应 Controller 类(Customer.cs)的代码
public ActionResult CreateCustomerData()
{
return PartialView("_CreateCustomer");
}
[HttpPost]
public ActionResult CreateCustomer(Models.Customer model)
{
if (ModelState.IsValid)
{
string firstName = model.FirstName;
int pinCode = model.Pincode;
return Content("Customer first name is " + firstName);
}
else
{
return RedirectToAction("index");
}
}
最佳答案
当您保存表单时，您的网站会发出如下所示的 POST 请求:
POST http://localhost:62186/Customer/CreateCustomer HTTP/1.1
FirstName=&LastName=&Address=&City=&State=&Pincode=&Phone=&Email=
当您的代码知道 Model.IsValid 为 false 时，它会使用 RedirectToAction 方法发送 302 HTTP 响应并重定向到/Customer 地址。然后浏览器像这样执行 GET 请求:
GET http://localhost:62186/Customer HTTP/1.1
如您所见，您的请求数据已完全丢失。您需要做的第一件事是将 RedirectToAction 替换为 View 方法并指向您的父 View 。它将在 POST 请求的上下文中加载您的 View 。这是您的 POST 操作的样子:
[HttpPost]
public ActionResult CreateCustomer(Models.Customer model)
{
if (ModelState.IsValid)
{
string firstName = model.FirstName;
int pinCode = model.Pincode;
return Content("Customer first name is " + firstName);
}
else
{
return View("Index");
}
}
当您的 View 在 POST 请求后呈现并且您单击“添加新客户”时，将执行另一个 GET 请求并且参数丢失:
GET http://localhost:62186/Customer/CreateCustomerData HTTP/1.1
避免此错误的最佳方法是在父 View 中预呈现局部 View 。您可以使用 Html.Partial 方法将局部 View 加载到另一个 View 中。然后你可以用 div 元素围绕这个声明并使用 JavaScript 来隐藏和显示它。您的父 View 应如下所示:
@{
ViewBag.Title = "CustomerMain";
Layout = "~/Views/Shared/_Layout.cshtml";
}
@Scripts.Render("~/bundles/jquery")
<html>
<head>
<script type="text/javascript">
$(document).ready(function () {
$("#CusForm").hide();
$('#CusCreatebtn').click(function (event) {
$("#CusForm").show();
});
});
</script>
</head>
<body>
<div id="CusDynamicPortion">
<div id="CusForm">
@Html.Partial("_CreateCustomer")
</div>
<table id="customeroptions" style="width:100%">
<tr>
<td><input type="button" id="CusCreatebtn" value="Add New Customer" /> </td>
<td><input type="button" id="CusViewbtn" value="View Customer" /> </td>
</tr>
</table>
</div>
</body>
</html>
通过这两个修复，您的代码将正常工作。但是如果你想使用 AJAX 来加载和发送表单，那么你应该阅读一些关于 Ajax.BeginForm 辅助方法的内容。
关于c# - 在 ASP.NET MVC 中，当 Model.isvalid 为 false 时如何将验证消息重定向并显示回局部 View ，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/33256403/