动态MVC路由：动作名称-Java 学习之路

我有控制器名称：关于和操作名称：索引 . 但我希望URL像这样（动作名称将动态）

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

Routing

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

Controller

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

        public ActionResult Index(string name)
        {

            return View();
        }

    }
}

View

@{
    ViewBag.Title = "Index";
}

<h2>Index About</h2>

3 回答

这应该工作 .

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

回复于 2024-04-25T23:45:23+08:00

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

回复于 2024-04-25T23:45:23+08:00

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)

回复于 2024-04-25T23:45:23+08:00

动态MVC路由：动作名称

3 回答

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