每当我尝试使用模型将视图中的值传递给控制器时,它在控制器中显示为null。我在项目的另一部分尝试了非常相似的东西并没有遇到这个问题。为什么这会回来?
以下是控制器的代码:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using MovieProject2.Models;
namespace MovieProject2.Controllers
{
public class ReviewsController : Controller
{
public ActionResult Edit(int id = -1)
{
if (id < 0)
return HttpNotFound();
MovieReview review = MovieADO.getReviewByID(id);
return View(review);
}
[HttpPost]
public ActionResult Edit(MovieReview review)
{
if (review == null) return HttpNotFound();
return View(review);
}
}
查看:
@model MovieProject2.Models.MovieReview
@{
ViewBag.Title = "Edit Review";
}
@{ //Not null here
if(@Model != null) {<h2>Edit Review for @Model.MovieReviewed.Title</h2>
<h4>Reviewed by @Model.Reviewer.Username</h4>}
else{<h2>Not Found</h2>
}
}
@using (Html.BeginForm())
{
Html.ValidationSummary(true);
<fieldset>
<div class="editor-label">
@Html.LabelFor(itemModel => Model.Rating)
</div>
<div class="editor-label">
@Html.EditorFor(itemModel => Model.Rating)
@Html.ValidationMessageFor(itemModel => Model.Rating)
</div>
<div class="editor-label">
@Html.LabelFor(itemModel => Model.Review)
</div>
<div class="editor-label">
@Html.EditorFor(itemModel => Model.Review)
@Html.ValidationMessageFor(itemModel => Model.Review)
</div>
<p>
<input type="submit" value="Change" />
</p>
</fieldset>
}
型号:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
namespace MovieProject2.Models
{
public class MovieReview
{
public int ReviewID { get; set; }
public Movie MovieReviewed { get; set; }
public User Reviewer { get; set; }
public int Rating { get; set; }
public string Review { get; set; }
public DateTime DateReviewed { get; set; }
public MovieReview() { }
}
}
答案 0 :(得分:5)
而不是
[HttpPost]
public ActionResult Edit(MovieReview review)
写
[HttpPost]
public ActionResult Edit(MovieReview model)
(并在该方法中将其从评论重命名为模型。它应该有效。
OR
将MovieReview.Review的属性重命名为其他内容(例如,Review1)。您不能为属性使用相同的名称并传递模型对象(不区分大小写)
答案 1 :(得分:0)
你必须尝试这个
@using (Html.BeginForm("Action methodName","controllerName"))