umbraco 7.1 logout (razor syntax C#)

Question

I have an umbraco 7.1 website and I use its own membership system.I can easily log in,see my status,even change password BUT there is no logout command and I don't know what method I should use for umbraco 7 to logout current user.I know i can clear the cookie and member identity but it always gives a runtime error. Help me! :D

Answer 1

In the backoffice (Admin) you can add a new Partial View and choose the Login Status template and you will get the following code.

@inherits Umbraco.Web.Mvc.UmbracoTemplatePage

@using System.Web.Mvc.Html
@using ClientDependency.Core.Mvc
@using Umbraco.Web
@using Umbraco.Web.Models
@using Umbraco.Web.Controllers

@{
    var loginStatusModel = Members.GetCurrentLoginStatus();

    Html.EnableClientValidation();
    Html.EnableUnobtrusiveJavaScript();
    Html.RequiresJs("/umbraco_client/ui/jquery.js");
    Html.RequiresJs("/umbraco_client/Application/JQuery/jquery.validate.min.js");
    Html.RequiresJs("/umbraco_client/Application/JQuery/jquery.validate.unobtrusive.min.js");

    var logoutModel = new PostRedirectModel();


    //Here you can specify a redirect URL for after logging out, by default umbraco will simply
    //redirect to the current page. Example to redirect to the home page:

    //logoutModel.RedirectUrl = "/"; 

}   // NOTE: This RenderJsHere code should be put on your main template page where the rest   of your script tags are placed
@Html.RenderJsHere()
@if (loginStatusModel.IsLoggedIn)
{
    <p>You are currently logged in as @loginStatusModel.Name</p>

   using (Html.BeginUmbracoForm<UmbLoginStatusController>("HandleLogout"))
   {
        <fieldset>
            <legend>Logout</legend>
            <button>Logout</button>
        </fieldset>

       @Html.HiddenFor(m => logoutModel.RedirectUrl)
   }
}