asp.net-mvc asp.net mvc - 将部分数据模型传递给部分视图

asp.net mvc - pass partial data model to partial view

提问by Nir

I wish to build a partial view that gets a model column and print it. Something like that:

我希望构建一个获取模型列并打印它的局部视图。类似的东西：

At the view:

在视图中：

@model IEnumerable<products_comparison.Models.Product>
@{
ViewBag.Title = "Index";

var Brand = (from r in Model
             select r.Brand).Distinct();
}
<h2>
Index</h2>

@Html.RenderPartial("_DisplayAttribute",Brand)

And at the partial view:

在局部视图中：

<table>
    <tr>
        <th>
            Brand
        </th>
    </tr>
    @foreach (var row in Model)
    {
        <tr>
            <td>
                @Html.DisplayFor(r => row)
            </td>
        </tr>
    }
</table>

There are a few problems I run into:

我遇到了几个问题：

1. The compiler doesnt allow me to send Barnd to the partial view.
2. If you look at the partial view code you will see the word Brand, which is the column name. I dont wish to hard-coded the word "Brand" in the partial view, instead I like that the column name will be there.
3. In the partial view I need to put @model products_comparison.Models.Product, but I dont want to send the hole table. I want to send only one column - But I dont know what to put there..

1. 编译器不允许我将 Barnd 发送到局部视图。
2. 如果您查看部分视图代码，您将看到 Brand 一词，即列名。我不希望在局部视图中硬编码“品牌”这个词，相反我喜欢列名将在那里。
3. 在局部视图中我需要放置@model products_comparison.Models.Product，但我不想发送孔表。我只想发送一栏 - 但我不知道该放什么..

Thanks!

谢谢！

EDIT:

编辑：

Just to clear one thing, I want that the view will call the same partial view for each column in the table(for most of the columns in the table anyway) and each time I'll send a different column(distinct value column to be exact).

只是为了清除一件事，我希望视图将为表中的每一列调用相同的局部视图（无论如何对于表中的大多数列），并且每次我都会发送一个不同的列（不同的值列是精确的）。

回答by Darin Dimitrov

Start by refactoring and putting the right logic into the right place. This LINQ query has strictly nothing to do in a view. A view is not supposed to do any LINQ queries or whatever to pull data. A view is supposed to work with data that it is passed to it from the controller action under the form of a view model. A controller action builds and passes an adapted view model that you define for the view.

首先重构并将正确的逻辑放在正确的位置。这个 LINQ 查询与视图无关。视图不应该执行任何 LINQ 查询或任何提取数据的操作。视图应该以视图模型的形式处理从控制器操作传递给它的数据。控制器操作构建并传递您为视图定义的适配视图模型。

So as always you start by defining a view model that will be adapted to the requirements of your view:

因此，与往常一样，您首先定义一个将适应您的视图要求的视图模型：

public class MyViewModel
{
    public IEnumerable<Brand> Brands { get; set; } 
}

then you write a controller action that will populate this view model and pass it to the view:

然后你编写一个控制器动作来填充这个视图模型并将它传递给视图：

public ActionResult Foo()
{
    IEnumerable<products_comparison.Models.Product> products = ...
    var model = new MyViewModel
    {
        Brands = (from r in Model select r.Brand).Distinct()
    };
    return View(model);
}

then a view:

然后是一个视图：

@model MyViewModel
<table>
    <tr>
        <th>
            Brand
        </th>
    </tr>
    @Html.DisplayFor(x => x.Brands)
</table>

and finally you could define a corresponding display template which will automatically be rendered for each element of the Brandscollection of your view model (~/Views/Shared/DisplayTemplates/Brand.cshtml):

最后，您可以定义一个相应的显示模板，该模板将自动Brands为您的视图模型 ( ~/Views/Shared/DisplayTemplates/Brand.cshtml)集合的每个元素呈现：

@model Brand
<tr>
    <td>
        @Html.DisplayForModel()
    </td>
</tr>

回答by Alexandros B

For 1try changing @Html.RenderPartial("_DisplayAttribute",Brand)to @Html.Partial("_DisplayAttribute",Brand)

对于1尝试更改@Html.RenderPartial("_DisplayAttribute",Brand)为@Html.Partial("_DisplayAttribute",Brand)

You will also need to specify the model in the partial view like @model products_comparison.Models.Brandor something like it

您还需要在局部视图中指定模型，例如@model products_comparison.Models.Brand或类似的东西

Also please clarify 2 & 3 as they are not clear what you want

另外请澄清 2 和 3，因为他们不清楚你想要什么