在ASP.NET MVC中使用C#将模型作为参数从POST方法传递到同一控制器的方法时，模型获取NULL

@model YourNamespace.YourModel

<form action="/YourController/YourMethod" method="post">
    <input type="text" name="PropertyName" />
    <button type="submit">Submit</button>
</form>

[HttpPost]
public ActionResult YourMethod(YourModel model)
{
    if (ModelState.IsValid)
    {
        // 处理模型
    }
    return View();
}

public class YourModel
{
    public string PropertyName { get; set; }
}

@using (Html.BeginForm("YourMethod", "YourController", FormMethod.Post))
{
    @Html.TextBoxFor(m => m.PropertyName)
    <button type="submit">Submit</button>
}

public class UserModel
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
}

public class UserController : Controller
{
    [HttpGet]
    public ActionResult Index()
    {
        return View();
    }

    [HttpPost]
    public ActionResult Index(UserModel userModel)
    {
        if (ModelState.IsValid)
        {
            // 处理用户模型
            TempData["Message"] = $"Hello, {userModel.FirstName} {userModel.LastName}!";
        }
        else
        {
            TempData["Error"] = "Please fill out all fields.";
        }
        return RedirectToAction("Index");
    }
}

@model YourNamespace.UserModel

@using (Html.BeginForm("Index", "User", FormMethod.Post))
{
    @Html.TextBoxFor(m => m.FirstName)
    @Html.TextBoxFor(m => m.LastName)
    <button type="submit">Submit</button>
}

@if (TempData["Message"] != null)
{
    <p>@TempData["Message"]</p>
}

@if (TempData["Error"] != null)
{
    <p style="color:red;">@TempData["Error"]</p>
}