在Layout.cshtml中将ViewModel传递给Partial View调用 [英] Passing a ViewModel to a Partial View call in the Layout.cshtml
问题描述
背景
我正在尝试在正在使用的MVC应用程序上使用仪表板模板.
I am trying to make use of Dashboard template on an MVC application I'm working on.
我不确定如何将ViewModel传递给我称为Header的顶部栏.
I am not sure how I can pass a ViewModel to the top bar I've called Header.
代码
在我的_Layout.cshtml中，我像这样分割了HTML:
In my _Layout.cshtml, I have split the HTML like so:
<body class="">
@Html.Partial("_Header")
<div class="page-container row-fluid">
@Html.Partial("_Sidebar")
<div class="page-content">
<div class="content">
@RenderBody()
</div>
</div>
</div>
</body>
我猜这是错误的，因为现在我无法将ViewModel传递给标题部分，或者如果可以的话，我们不应该这样做吗?
I am guessing this is wrong because now I cannot pass a ViewModel to the header section, or if I can, we're not supposed to?
将其拆分的正确方法是什么?
What is the right way of splitting this?
推荐答案
使用@Html.Action()或@{ Html.RenderAction(); }代替@Html.Partial()来调用[ChildActionOnly]控制器方法，该方法将为仪表板初始化模型并返回部分信息.视图，例如
Instead of @Html.Partial(), use @Html.Action() or @{ Html.RenderAction(); } to call a [ChildActionOnly] controller methods that initializes your model for the dashboard and returns a partial view, for example
[ChildActionOnly]
public PartialViewResult Header()
{
// initialize a model
return PartialView("_Header", model)
}
并在布局中
@{ Html.RenderAction("Header", yourControllerName); }
另一种选择是，您在使用该布局的每个视图中使用的模型将需要一个属性，该属性是用于生成仪表板然后使用@Html.Partial("_Header", Model.yourDashBoardProperty)
The alternative is that the model you use in each view using that layout would need a property which is the model used to generate your dashboard and then use @Html.Partial("_Header", Model.yourDashBoardProperty)
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