C# 在 View MVC 中访问控制器变量
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/18488667/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Accessing Controller variable in View MVC
提问by Inkey
I have started to use MVC and i have set up a simple project that has a controller than returns a value from entity FrameWork. I have a controller for the index page which visual studio setup as default in its template.
我已经开始使用 MVC 并且我已经建立了一个简单的项目，它有一个控制器而不是从实体 FrameWork 返回一个值。我有一个用于索引页面的控制器，Visual Studio 将其设置为模板中的默认值。
I am hoping that this value will be returned to the index page of the website.
我希望这个值会返回到网站的索引页面。
Controller code
控制器代码
MISEntities db = new MISEntities();
public ActionResult Index()
{
ViewBag.Message = "Real Time Production";
var scrap = from r in db.tbl_dppIT
select r.CastGood;
return View(scrap);
}
How do i access the var scrap using razor? I have looked at the viewbag method and other razor syntax but cant seem to find how i access this item.
如何使用剃刀访问 var 废料？我查看了 viewbag 方法和其他 razor 语法，但似乎无法找到我如何访问此项目。
采纳答案by fourpastmidnight
Just because the controller variable is declared using the varkeyword doesn't mean it doesn't have a type. Rather, this tells C# that the type should be inferred by its context (in this case, an IEnumerable<T>of whatever type tbl_dppIT.CastGoodis).
仅仅因为使用var关键字声明控制器变量并不意味着它没有类型。相反，这告诉 C# 类型应该由其上下文推断（在这种情况下，IEnumerable<T>无论类型tbl_dppIT.CastGood是什么）。
So, the other three answers are partly correct, you need to define the type of model being passed into the View via the razor keyword @modelas noted by the other answers:
因此，其他三个答案部分正确，您需要定义通过 razor 关键字传递到 View 的模型类型，@model如其他答案所述：
// In your view Razor:
@model IEnumerable</* type of CastGood */>
There is an alternative to the three answers already specified, and that is sticking the scrapvariable into the ViewBagin your controller:
已经指定的三个答案还有一个替代方案，那就是将scrap变量粘贴到ViewBag控制器中：
ViewBag.Scrap = scrap;
When you access the expando property Scrapfrom the ViewBagin the view, it will be a dynamic object, so you won't get the aid of IntelliSense. But, it is another way, just so that you know all the possibilities.
当您Scrap从ViewBag视图中访问 expando 属性时，它将是一个动态对象，因此您将无法获得 IntelliSense 的帮助。但是，这是另一种方式，只是为了让您知道所有的可能性。
UPDATE:
更新：
Based on your comments below, it looks like if CastGood is a database column that is allowed to be null and whose type is int, then you'd want:
根据您在下面的评论，看起来如果 CastGood 是一个允许为 null 且其类型为 的数据库列int，那么您需要：
@model IEnumerable<int?>
HTH.
哈。
回答by Dustin Kingen
You need to declare the type of the variable scrap as the model inside the View using the @modelkeyword.
您需要使用@model关键字将变量 scrap 的类型声明为 View 内的模型。
// View.cshtml
@model IEnumerable<Namespace.ScrapType>
@foreach(var item in Model) {
<p>@item.SomeProperty</p>
}
See Part 3: Views and ViewModelsof the MVC Music Store example.
回答by Ufuk Hac?o?ullar?
You need to strongly type your view and it will available as the Model instance.
您需要强输入您的视图，它将作为模型实例使用。
@model IEumerable<MyType>
@foreach(var item in Model)
{
@*do something with it *@
}
