handlelogout gives me a ysod

Bear 40 posts 129 karma points

May 23, 2014 @ 17:46

"HandleLogout" gives me a YSOD

Can anyone tell me why I'm getting a YSOD from the following code? It's only happening when I click the "Logout" button.

@inherits Umbraco.Web.Mvc.UmbracoTemplatePage

@using Umbraco.Web
@using Umbraco.Web.Controllers

@if (Members.IsLoggedIn())
{
    <p>You are currently logged in as <strong>@HttpContext.Current.User.Identity.Name</strong></p>

    using (Html.BeginUmbracoForm<UmbLoginStatusController>("HandleLogout"))
    {
    <fieldset>
        <legend>Logout</legend>
        <button>Logout</button>
    </fieldset>
    }   
}
else
{
    <p>You are not logged in.</p>
}

Clicking the "Logout" button

[NullReferenceException: Object reference not set to an instance of an object.]
   Umbraco.Web.Controllers.UmbLoginStatusController.HandleLogout(PostRedirectModel model) +130
   lambda_method(Closure , ControllerBase , Object[] ) +93
   System.Web.Mvc.ReflectedActionDescriptor.Execute(ControllerContext controllerContext, IDictionary`2 parameters) +260
   System.Web.Mvc.ControllerActionInvoker.InvokeActionMethod(ControllerContext controllerContext, ActionDescriptor actionDescriptor, IDictionary`2 parameters) +38
   System.Web.Mvc.<>c__DisplayClass13.<InvokeActionMethodWithFilters>b__10() +123
   System.Web.Mvc.ControllerActionInvoker.InvokeActionMethodFilter(IActionFilter filter, ActionExecutingContext preContext, Func`1 continuation) +855266
   System.Web.Mvc.ControllerActionInvoker.InvokeActionMethodFilter(IActionFilter filter, ActionExecutingContext preContext, Func`1 continuation) +855266
   System.Web.Mvc.ControllerActionInvoker.InvokeActionMethodFilter(IActionFilter filter, ActionExecutingContext preContext, Func`1 continuation) +855266
   System.Web.Mvc.ControllerActionInvoker.InvokeActionMethodFilter(IActionFilter filter, ActionExecutingContext preContext, Func`1 continuation) +855266
   System.Web.Mvc.ControllerActionInvoker.InvokeActionMethodWithFilters(ControllerContext controllerContext, IList`1 filters, ActionDescriptor actionDescriptor, IDictionary`2 parameters) +309
   System.Web.Mvc.ControllerActionInvoker.InvokeAction(ControllerContext controllerContext, String actionName) +855200
   System.Web.Mvc.<>c__DisplayClass1d.<BeginExecuteCore>b__19() +40
   System.Web.Mvc.Async.<>c__DisplayClass1.<MakeVoidDelegate>b__0() +15
   System.Web.Mvc.Controller.EndExecuteCore(IAsyncResult asyncResult) +65
   System.Web.Mvc.Async.<>c__DisplayClass4.<MakeVoidDelegate>b__3(IAsyncResult ar) +15
   System.Web.Mvc.Controller.EndExecute(IAsyncResult asyncResult) +51
   System.Web.Mvc.<>c__DisplayClass8.<BeginProcessRequest>b__3(IAsyncResult asyncResult) +42
   System.Web.Mvc.Async.<>c__DisplayClass4.<MakeVoidDelegate>b__3(IAsyncResult ar) +15
   System.Web.Mvc.MvcHandler.EndProcessRequest(IAsyncResult asyncResult) +51
   System.Web.CallHandlerExecutionStep.System.Web.HttpApplication.IExecutionStep.Execute() +606

System.Web.HttpApplication.ExecuteStep(IExecutionStep step, Boolean& completedSynchronously) +288

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Jan Skovgaard 11280 posts 23678 karma points MVP 11x admin c-trib

May 24, 2014 @ 11:48

Hi Bear

What does your code for logging members out look like?

And what version of Umbraco are you using?

/Jan

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Bear 40 posts 129 karma points

May 24, 2014 @ 12:03

Hi Jan,

7.1.3 - I'm using the "snippets" from the macro section, so it's using the UmbLoginStatusController which is already in Core. That means it's using this code;

https://github.com/umbraco/Umbraco-CMS/blob/7.1.4/src/Umbraco.Web/Controllers/UmbLoginStatusController.cs

I'm very aware of how to roll my own, but I wanted to see if I could just get this working "out-of-the-box" so a designer could just cut and paste.

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Jan Skovgaard 11280 posts 23678 karma points MVP 11x admin c-trib

May 24, 2014 @ 15:54

Hi Bear

Aaah ok - Hmm, not sure wether that's actually a bug or not to be honest. Have not yet had the time to play around with these myself yet.

I guess it could not hurt to file it on the issue tracker

/Jan

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Lennart Stoop 304 posts 842 karma points

Jun 11, 2014 @ 15:33

Hi guys,

I just ran into the same problem, and it seems the model is not bound (hence the null reference exception).

As a simple workaround you can just create a PostRedirectModel in the view, and make sure it is bound to the form context.

@{

var logoutModel = new PostRedirectModel();
logoutModel.RedirectUrl = "/";

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");
}

@using (Html.BeginUmbracoForm<UmbLoginStatusController>("HandleLogout"))
{
    @Html.HiddenFor(m => logoutModel.RedirectUrl)
    <button>Logout</button>
}

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"HandleLogout" gives me a YSOD