这个问题已经在这里有了答案：            >            What is a NullReferenceException, and how do I fix it?                                    31个
因此,我正在创建一个强类型的视图.我的模型称为RestaurantReview.cs,如下所示：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;

namespace OdeToFood.Models
{
    public class RestaurantReview
    {
        public int Id { get; set; }
        public string Name { get; set; }
        public string City { get; set; }
        public string Country { get; set; }
        public int Rating { get; set; }
    }
}

我让Visual Studio基于此创建一个强类型的List模型,如下所示：

@model IEnumerable<OdeToFood.Models.RestaurantReview>

@{
    ViewBag.Title = "Index";
}

<h2>Index</h2>

<p>
    @Html.ActionLink("Create New", "Create")
</p>
<table>
    <tr>
        <th>
            @Html.DisplayNameFor(model => model.Name)
        </th>
        <th>
            @Html.DisplayNameFor(model => model.City)
        </th>
        <th>
            @Html.DisplayNameFor(model => model.Country)
        </th>
        <th>
            @Html.DisplayNameFor(model => model.Rating)
        </th>
        <th></th>
    </tr>

@foreach (var item in Model) {
    <tr>
        <td>
            @Html.DisplayFor(modelItem => item.Name)
        </td>
        <td>
            @Html.DisplayFor(modelItem => item.City)
        </td>
        <td>
            @Html.DisplayFor(modelItem => item.Country)
        </td>
        <td>
            @Html.DisplayFor(modelItem => item.Rating)
        </td>
        <td>
            @Html.ActionLink("Edit", "Edit", new { id=item.Id }) |
            @Html.ActionLink("Details", "Details", new { id=item.Id }) |
            @Html.ActionLink("Delete", "Delete", new { id=item.Id })
        </td>
    </tr>
}

</table>

运行网站时,在“ @foreach(模型中的变量项)”行中出现空指针异常,突出显示模型对象,并指出“对象引用未设置为对象的实例”.

因为我什至没有编写代码,所以我不是很了解这段代码是怎么错的,Visual Studio做到了.这是怎么回事

解决方法:

您的控制器显示已通过RestaurantReview IEnumerable.例：

public class HomeController : Controller { //suppose this is your Home
    public ActionResult Index() {
        IEnumerable<OdeToFood.Models.RestaurantReview> model;
        model = from m in db.RestaurantReviews
                ... //your query here
                select m;
        return View(model); //pass the model here
    }

那么您将不会获得null异常