MVC4复合型模式是后后空 [英] MVC4 Complex Type Model is null after post

问题描述

这是我的模型

public class AdministrationModel
{
  public string FirstName { get; set; }
  public string LastName { get; set; }
  public string EmailAddress { get; set; }
  public bool IsApproved { get; set; }
}

这是我的控制器

public ActionResult GetTabContent(string id)
{
  switch (id)
  {
   case "tab3":
   model = GetAllUsersInfo();
   viewName = "Administration";
   break;
   }
   return View(viewName);
 }

  private List<AdministrationModel> GetAllUsersInfo()
  {
    List<AdministrationModel> userList = new List<AdministrationModel>();
    foreach (MembershipUser user in Membership.GetAllUsers())
    {
      UserProfile userProfile = UserProfile.GetUserProfile(user.UserName);
      userList.Add(new AdministrationModel { EmailAddress = user.Email,                       IsApproved = user.IsApproved, FirstName = userProfile.FirstName, LastName = userProfile.LastName });
    }

    return userList;
  }

这是我的看法。

@model List<AdminContainerModel>
@using (Html.BeginForm("Administration", "Account"))
{
  <fieldset>
    <div>
      @foreach (AdministrationModel AM in Model)
      {
        <div>
         <div class="colFull">@Html.DisplayFor(modelItem => AM.FirstName)</div>
         <div class="colFull">@Html.DisplayFor(modelItem => AM.LastName)</div>
         <div class="colFull">@Html.DisplayFor(modelItem => AM.EmailAddress)</div>
         <div class="colPartial"><input type="checkbox" checked="@AM.IsApproved"/>            </div>
      <div class="clear"></div>
    </div>
  }
</div>
 <input type="submit" value="Update Account" />
 </fieldset>
}

当用户点击更新帐户按钮它关系到控制器

When the user clicks the Update Account button it goes to the controller

  [HttpPost]
  public ActionResult Administration(List<AdministrationModel> model)
  {
     return View();
  }

这个方法中，模式始终为空。然而，呈现一切都是完美的，并显示我想让它显示什么视图。我在做什么错了？

inside this method, model is always null. however the View that renders everything is perfect and shows what I want it to show. What am I doing wrong?

推荐答案

在使用集合，以便正确地处理它们，所以它们是模型绑定上后没有任何额外的腿部的工作，你需要确保他们正确索引，您可以通过使用一个循环做到这一点，是这样的：

When using collections, in order to correctly process them so they are model-bound on post without any additional leg work, you need to make sure they are indexed correctly, you can do this by using a for loop, something like:

@for (int i = 0; i < Model.Count; i++)
{
    @Html.HiddenFor(m => m[i].FirstName)
    @Html.HiddenFor(m => m[i].LastName)
    @Html.HiddenFor(m => m[i].EmailAddress)
    <div>
        <div class="colFull">@Html.DisplayFor(m => m[i].FirstName)</div>
        <div class="colFull">@Html.DisplayFor(m => m[i].LastName)</div>
        <div class="colFull">@Html.DisplayFor(m => m[i].EmailAddress)</div>
        <div class="colPartial">@Html.CheckBoxFor(m => m[i].IsApproved)</div>
        <div class="clear"></div>
    </div>
}

这应该没有任何其他code型绑定：）

That should model bind without any other code :)

编辑：对不起我忘了，默认情况下displayFors不要把对模型绑定正确的属性，增加hiddenFors对于不具有editorFor其他领域

EDIT2：根据在评论你的另一个问题是，如果是面向公众的，你不希望他们更改任何隐藏的使用开发工具价值，请尝试以下操作：

好了，所以你不希望他们改变hiddenFors，这很好，但你会需要某种形式的ID，所以你知道哪些客户是当数据发布，我建议，而不是在上面有这些code：

Ok so you don't want them to change the hiddenFors, that's fine, but you will need some sort of ID so you know which client is which when the data is posted, I suggest that instead of having these in the above code:

@Html.HiddenFor(m => m[i].FirstName)
@Html.HiddenFor(m => m[i].LastName)
@Html.HiddenFor(m => m[i].EmailAddress)

替换为它们：

@Html.HiddenFor(m => m[i].ClientId)

你不回发的名字，姓氏或电子邮件地址，只是为了被选中，取消选中实际的客户端的引用的方式。

That way you're not posting back the firstname, lastname or email address, just a reference to the actual client that is ticked, unticked.

为了回答您的其他问题在关于跟踪原始值的，在你的控制器方法，你可以随便去从数据库中获取的原始值，那么这里是你如何能发现哪些是不同的评论，是这样的：

[HttpPost]
public ActionResult Administration(List<AdministrationModel> model)
{
    var originalMatches = GetAllUsersInfo();

    var differences = (from o in originalMatches
                      join c in model on o.ClientId equals c.ClientId
                      where c.IsApproved != o.IsApproved).ToList()

    return View();
}

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