部分视图中的 asp.net MVC3 @model

发布于 2024-12-10 01:35:48 字数 197 浏览 0 评论 0原文

在我的控制器中，我进行如下初始化：

using mylib.product;
using mylib.factory;

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

如何在部分视图中使用 @model 关键字执行相同的操作？

谢谢

原文

In my controller I do initialization like this:

using mylib.product;
using mylib.factory;

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

How do I do the same thing using the @model keyword in a partial view?

Thanks

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霊感 2024-12-17 01:35:48

如果您尝试将名称空间/类添加到视图中，那么它是：

@using mylib.product;

If you are trying to add namespaces/classes to you view, then it's:

@using mylib.product;

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聽兲甴掵 2024-12-17 01:35:48

将模型解析为视图；

return View("ViewName");

我应该通过视图

@model Project.Namespace.Class

I should parse the model to the view by

return View("ViewName");

and in the view;

@model Project.Namespace.Class

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孤君无依 2024-12-17 01:35:48

您应该使用视图模型：

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

它将从控制器操作传递到视图：

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

然后您的视图将被强类型化到该视图模型：

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

You should use view models:

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

which will be passed to the view from the controller action:

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

and then your view will be strongly typed to this view model:

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

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深爱不及久伴 2024-12-17 01:35:48

我认为你需要将不同类的多个实例传输到View。（我是对的吗？）如果是，我建议使用ViewBag。像这样的事情：

// 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)

不要忘记 ViewBag 是一个动态容器，当您想在 View 中使用它的值时，您需要将其转换为类型

I think you need to transfer more than one instance of different classes to View.(Am I right?) If yes, I suggest to use ViewBag for it. Something like this:

// 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)

Do not forgot that ViewBag is a Dynamic container and you need to Cast it to a type when you want to use its value in View

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