How to return different view, but presererve ViewModel in OnActionExecuting

Question

Hi!

I'm trying to return a different view if a condition is met. I want to preserve the Model passed into the view from the action.

    protected override void OnActionExecuting(ActionExecutingContext filterContext)
    {
        var subAction = filterContext.RequestContext.RouteData.Values["subaction"].ToString();
        var action = filterContext.RequestContext.RouteData.Values["action"].ToString();

        if (!string.IsNullOrEmpty(subAction))
        {
            var view = (ViewResultBase) (filterContext.Result);
            filterContext.Result = View(action + subAction, view.ViewData.Model);
        }

        base.OnActionExecuting(filterContext);
    }

When doing it this way I get an "Object reference not set to an instance of an object.". Obviously, the Model haven't been set?

The reason I'm doing it this way, is that I want to keep the naming of the view as simple as possible. And my URL looks like this: /Global/Modules/Admin/Users/Create. That would return the view "UsersCreate". Which works. But the Model is either empty or null!

UPDATE

Actually. It just hit me. The behaviour is correct. Because I'm only returning a View. The action is never executed. The Users-action gets executed, but returns the view UsersCreate. How can I do a RedirectToAction kinda thing, without actually redirect.

Answer 1

OnActionExecuting is run before the action so no model exists yet. You may try with OnActionExecuted which is called once the action returns an ActionResult.