如何将数据传递到布局中的 PartialView？

发布于 2024-12-07 12:12:43 字数 186 浏览 0 评论 0原文

我有一个 _layout.cshtml 包含这一行：

@{Html.RenderPartial("Menu");}

现在我想将模型传递到此 RenderPartial 函数中。该模型可以从我的存储库中读取。

如何以及在哪里（在代码中）可以完成此操作？

谢谢！

原文

I have a _layout.cshtml containing this row:

@{Html.RenderPartial("Menu");}

Now I want to pass in a model into this RenderPartial-function. This model can be read from my repository.

How and where(in code) can this be done?

Thanks!

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情场扛把子 2024-12-14 12:12:43

RenderPartial 有一个重载，可以将对象发送到局部视图。
不要忘记在局部视图的顶部定义 @model 以使用正确的对象类型。

@Html.RenderPartial("ViewName",object)

额外信息： MSDN

评论后编辑：

我认为创建一个接受存储库的 MenuController 会更容易。然后让它创建一个视图，该视图将所需的存储库作为模型，然后使用 foreach 将每个菜单项渲染为操作链接，并将菜单信息传递给它。

所以你会在你的 _layout.cshtml 中包含这个：

<div id="Menu">
    @{Html.RenderAction("Menu", "Menu");}
</div>

在你的 MenuController 中：这个

public class MenuController : Controller
{
    private IMenuRepository _repository;

    public MenuController(IMenuRepository repo)
    {
        _repository = repo;
    }
    //
    // GET: /Menu/

    public PartialViewResult Menu(string menu = null)
    {
        ViewBag.SelectedMenu = menu;

        IEnumerable<MenuInfoObject> menus= _repository.Menus;
        return PartialView(menus);
    }
}

在你的 MenuView 中：

    @model IEnumerable<MenuInfoObject>
@{
    Layout = null;
}
@foreach (var item in Model)
{
    @Html.RouteLink(item.MenuName, new
{
    controller = item.ControllerInfo,
    action = item.ActionInfo,
}, new
       {
           @class = item.Menu == ViewBag.SelectedMenu ? "selected" : null
       })
}

这会更接近解决方案吗？

RenderPartial has an overload that can take an object to send it to the partial view.
Don't forget to define your @model at the top of your partialview to work with the right object type.

@Html.RenderPartial("ViewName",object)

Extra info: MSDN

Edit after comment:

I think it would be easier to create a MenuController that takes in the repository. Then let it create a view which takes in the required repository as it's model, then with a foreach render every menu-item as actionlinks, passing the menu info to it.

So you would have this in your _layout.cshtml:

<div id="Menu">
    @{Html.RenderAction("Menu", "Menu");}
</div>

This in your MenuController:

public class MenuController : Controller
{
    private IMenuRepository _repository;

    public MenuController(IMenuRepository repo)
    {
        _repository = repo;
    }
    //
    // GET: /Menu/

    public PartialViewResult Menu(string menu = null)
    {
        ViewBag.SelectedMenu = menu;

        IEnumerable<MenuInfoObject> menus= _repository.Menus;
        return PartialView(menus);
    }
}

And your MenuView:

    @model IEnumerable<MenuInfoObject>
@{
    Layout = null;
}
@foreach (var item in Model)
{
    @Html.RouteLink(item.MenuName, new
{
    controller = item.ControllerInfo,
    action = item.ActionInfo,
}, new
       {
           @class = item.Menu == ViewBag.SelectedMenu ? "selected" : null
       })
}

Would that be any closer to a solution?

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放血 2024-12-14 12:12:43

还有另一种解决方案将数据传递到布局中的部分视图。您可以简单地将其添加到 _Layout.cshtml 文件

@Html.Action("ActionName","ControllerName")

和 Controller 中：

    [ChildActionOnly]
    public ActionResult ActionName()
    {
        var model = new YourModel();
        return PartialView(model);

    }

ChildActionOnly 属性确保操作方法只能作为子方法调用。此操作将在布局中使用模型渲染相应的部分视图。

There is another solution as well to pass data to a partial view in Layout. You can simply add this in your _Layout.cshtml file

@Html.Action("ActionName","ControllerName")

And in your Controller :

    [ChildActionOnly]
    public ActionResult ActionName()
    {
        var model = new YourModel();
        return PartialView(model);

    }

The ChildActionOnly attribute ensures that an action method can be called only as a child method. This action will render the corresponding partial view with model in Layout.

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