ASP.Net MVC3 Menu with different Model as Content (partial view, renderpage?)
i have a page with a fix menu on the left side. This Partial View needs a different Model as the Main Page (Content).
Masterpage/Layout:
<body>
<div id="IndexMenu">
<div id="IndexMenuInne开发者_开发问答r">@RenderPage("~/Views/Admin/part/_Menu.cshtml", new { LocationAdminModelCollection = new Model; })</div>
</div>
<div id="BodyContent">
@RenderBody()
</div>
Index/Content Page which where called at start:
@model Survey.WebApplication.Models.ChecklistDetailsModel
@{
ViewBag.Title = "Survey Administration";
Layout = "~/Views/Admin/_Layout.cshtml";
}
<link href="@Url.Content("~/Content/Admin/Menu.css")" rel="stylesheet" type="text/css" />
<div id="IndexSubMenu">sub_Menu</div>
<div>
<div id="IndexMenuInner"></div>
</div>
My Menu:
@model Survey.WebApplication.Models.LocationAdminModelCollection
@{
Layout = null;
}
<div class="menuLocation">
</div>
Ho can i do this?
I would use Html.RenderAction to render an Action on your Controller. In that action, you simply create the Model that your Menu needs, and pass out the Menu.cshtml partial View as a PartialViewResult
So instead of @RenderPage("~/Views/Admin/part/_Menu.cshtml", new { LocationAdminModelCollection = new Model; })
you'd do:
@{ Html.RenderAction("Menu", "Site"); }
Where Site is your SiteController and Menu is something like:
public ActionResult Menu()
{
return PartialView("Menu", new { LocationAdminModelCollection = new Model });
}
Disclaimer
Code not tested :)
加载中,请稍侯......
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