MVc dynamic Menu using Partial View

Mark Tabor

I have made a simple Dynamic MVC menu application in which i have an entity model and i bind this entity model from database table {Menu table} , I have created a simple MVC application in which it automatically add home controller and i wrote the below c0de in home controller

private TestEntities testEntities = new TestEntities();

public ActionResult Index()

{

ViewBag.MenuLevel1 = this.testEntities.Menus.Where(menu=>menu.ParentId==null ).ToList();

return View();

}

Secondly I have Menu Folder Under Views and in this Menu folder i have index view and my index view is just look like that

@model Menu.Models.Menu

@foreach (var menuLevel1 in ViewBag.MenuLevel1)

{

@menuLevel1.Name

@if (menuLevel1.Child.Count > 0)

{

@foreach (var menuLevel2 in menuLevel1.Child)

{

@menuLevel2.Name

}

Now i want to make this Menu item as partial view and want to call it from _LAyout.cshtml how to do that ? i have check lot of example non of them makes me happy :)

2 Replies

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Mark Tabor7y ago
in my case my partial view get the data from the model so it is not just a partial view it is having model as well . i have tried
@{Html.RenderAction("MenuPartial", "Home");}
public ActionResult MenuPartial()
{
using (var db = new MenuModel())
{
var d = db.tblMainMenus.Include(c => c.tblSubMenus).ToList();
return PartialView("_MenuPartial",d);
}
}
I have _MenuPartial view in shared folder , btu it did not work
0
Bhuvanesh Mohankumar10y ago
Create a partial view and add the set of menu code to the partial view andcreate the view in the shared folder, which can be used any where in the applications.
Consume the partial view in the main viewwhere ever its required,
2. @Html.Partial("PartialViewExample")
@{
Html.RenderPartial("PartialViewExample");
}