是否可以将formcollection
转换为已知的“模型”?
[HttpPost]
public ActionResult Settings(FormCollection fc)
{
var model=(Student)fc; // Error: Can't convert type 'FormCollection' to 'Student'
}
注意:由于某些原因,我不能使用ViewModel。
这是我的代码VIEW:Settings.cshtml
@model MediaLibrarySetting
@{
ViewBag.Title = "Library Settings";
var extensions = (IQueryable<MediaLibrarySetting>)(ViewBag.Data);
}
@helper EntriForm(MediaLibrarySetting cmodel)
{
<form action='@Url.Action("Settings", "MediaLibrary")' id='MLS-@cmodel.MediaLibrarySettingID' method='post' style='min-width:170px' class="smart-form">
@Html.HiddenFor(model => cmodel.MediaLibrarySettingID)
<div class='input'>
<label>
New File Extension:@Html.TextBoxFor(model => cmodel.Extention, new { @class = "form-control style-0" })
</label>
<small>@Html.ValidationMessageFor(model => cmodel.Extention)</small>
</div>
<div>
<label class='checkbox'>
@Html.CheckBoxFor(model => cmodel.AllowUpload, new { @class = "style-0" })<i></i>
<span>Allow Upload.</span></label>
</div>
<div class='form-actions'>
<div class='row'>
<div class='col col-md-12'>
<button class='btn btn-primary btn-sm' type='submit'>SUBMIT</button>
</div>
</div>
</div>
</form>
}
<tbody>
@foreach (var item in extensions)
{
if (item != null)
{
<tr>
<td>
<label class="checkbox">
<input type="checkbox" value="@item.MediaLibrarySettingID"/><i></i>
</label>
</td>
<td>
<a href="javascript:void(0);" rel="popover" class="editable-click"
data-placement="right"
data-original-title="<i class='fa fa-fw fa-pencil'></i> File Extension"
data-content="@EntriForm(item).ToString().Replace("\"", "'")"
data-html="true">@item.Extention</a></td>
</tr>
}
}
</tbody>
控制器:
[HttpPost]
public ActionResult Settings(FormCollection fc)//MediaLibrarySetting cmodel - Works fine for cmodel
{
var model =(MediaLibrarySetting)(fc);// Error: Can't convert type 'FormCollection' to 'MediaLibrarySetting'
}
data-content
和data-
属性是bootstrap popover。
答案 0 :(得分:6)
MVC中的另一种方法是使用TryUpdateModel。
实施例: TryUpdateModel或UpdateModel将从已发布的表单集合中读取并尝试将其映射到您的类型。我发现这比手动手动映射字段更优雅。
[HttpPost]
public ActionResult Settings()
{
var model = new Student();
UpdateModel<Student>(model);
return View(model);
}
答案 1 :(得分:3)
好问题! 在制作通用基础控制器的过程中也是如此,模型独立。感谢很多人,最后一个是@GANI,已经完成了。
Type ViewModelType
在子类控制器中设置为您想要的任何内容。
public ActionResult EatEverything(FormCollection form)
{
var model = Activator.CreateInstance(ViewModelType);
Type modelType = model.GetType();
foreach (PropertyInfo propertyInfo in modelType.GetProperties())
{
var mykey = propertyInfo.Name;
if (propertyInfo.CanRead && form.AllKeys.Contains(mykey))
{
try
{
var value = form[mykey];
propertyInfo.SetValue(model, value);
}
catch
{
continue;
}
}
}
现在您从未知表单收到的所有内容都在您的真实模型中,您可以通过此帖https://stackoverflow.com/a/22051586/7149454
继续进行验证答案 2 :(得分:2)
你可以试试这种方式
public ActionResult Settings(FormCollection formValues)
{
var student= new Student();
student.Name = formValues["Name"];
student.Surname = formValues["Surname"];
student.CellNumber = formValues["CellNumber"];
return RedirectToAction("Index");
}