Question

我有控制器名称：关于和操作名称：索引。但我希望URL就像这样（动作名称将是动态的）

www.example.com/about/aaa www.example.com/about/bbb www.example.com/about/ccc

路由

  routes.MapRoute(
                name: "About",
                url: "{controller}/{name}",
                defaults: new { controller = "About", action = "Index"}

控制器

public class AboutController : Controller
    {
        //
        // GET: /About/

        public ActionResult Index(string name)
        {

            return View();
        }

    }
}

查看

@{
    ViewBag.Title = "Index";
}

<h2>Index About</h2>

Answer 1

这应该有用。

routes.MapRoute(
    name: "About",
    url: "About/{name}",
    defaults: new
    {
        controller = "About",
        action = "Index"
    });

确保您的默认路线存在，并在关于路线

之后

Answer 2

routes.MapRoute(
    name: "About",
    url: "about/{name}/{id}",
    defaults: new { controller = "About", action = "Index", id=UrlParameter.Optional}

Answer 3

您可以将ActionResult名称作为参数传递：

 public ActionResult Index(string name)
        {

            return View(name);
        }


public ActionResult First()
        {

            return View();
        }

public ActionResult Second()
        {

            return View();
        }

在视图中：

@Html.ActionLink("Get Action named Firts" "Index", "Home", new {name = "First"}, null)

动态MVC路由：动作名称

3 个答案: