Question

在我的控制者中,我这样做是初步的:

using mylib.product;
using mylib.factory;

product p = new product();
factory f = new factory(p);

从部分角度看,我如何用“@model”关键词做同样的事?

增编

Answer 1

我认为,你们需要将不止一个不同类别的例子转移给观点。如果是,我建议为此使用“观点”。与此类似:

// Controller
=========
product p = new product(); 
factory f = new factory(p);
....
// Add some value for p and f 
ViewBag.Product = p;
ViewBag.Factory = f;
return View();

// View
=========
var p = (product) ViewBag.Product;
var f = (factory) ViewBag.Factory;
// now you have access to p and f properties, for example:
@Html.Label(p.Name)
@Html.Label(f.Address)

不要放弃观点Bag是一个动态的集装箱,你需要把它放在一种类型上,因为你想在观点中使用其价值。

Answer 2

如果你试图在你看来增加姓名/舱位,那么:

@using mylib.product;

Answer 3

我应该把这一模式同大家一样看待。

return View("ViewName");

and in the view;

@model Project.Namespace.Class

Answer 4

你们应当使用观点模式:

public class MyViewModel
{
    public string Name { get; set; }
    public string Address { get; set; }
}

将由控制者行动考虑:

public ActionResult Index()
{
    product p = new product(); 
    factory f = new factory(p);   
    var model = new MyViewModel
    {
        Name = p.Name,
        Address = f.Address
    }
}

之后,您的意见将严格归纳为这一看法模式:

@model MyViewModel
@Html.DisplayFor(x => x.Name)
@Html.DisplayFor(x => x.Address)

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