18

In MVC 5, you can do something like this inside an IActionFilter, to check if an attribute has been declared on the the current action (or at controller scope)

public void OnActionExecuting(ActionExecutingContext filterContext)
{
    // Stolen from System.Web.Mvc.AuthorizeAttribute
    var isAttributeDefined = filterContext.ActionDescriptor.IsDefined(typeof(CustomAttribute), true) ||
                             filterContext.ActionDescriptor.ControllerDescriptor.IsDefined(typeof(CustomAttribute), true);

}

So if your controller defines the attribute like so, this works.

[CustomAttribute]
public ActionResult Everything()
{ .. }

Is it possible to do the same in ASP.NET Core MVC (inside an IActionFiler)?

Tseng
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Matt Roberts
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3 Answers3

23

Yes you can do it. Here is similar code for ASP.NET Core.

public void OnActionExecuting(ActionExecutingContext context)
{
    var controllerActionDescriptor = context.ActionDescriptor as ControllerActionDescriptor;
    if (controllerActionDescriptor != null)
    {
        var isDefined = controllerActionDescriptor.MethodInfo.GetCustomAttributes(inherit: true)
            .Any(a => a.GetType().Equals(typeof(CustomAttribute)));
    }
}
Anuraj
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7

If you need to check the attribute, not only for a method but also for the whole controller in .NET Core, here is how I did it:

var controllerActionDescriptor = actionContext.ActionDescriptor as ControllerActionDescriptor;

if (controllerActionDescriptor != null)
{
    // Check if the attribute exists on the action method
    if (controllerActionDescriptor.MethodInfo?.GetCustomAttributes(inherit: true)?.Any(a => a.GetType().Equals(typeof(CustomAttribute))) ?? false)
        return true;

    // Check if the attribute exists on the controller
    if (controllerActionDescriptor.ControllerTypeInfo?.GetCustomAttributes(typeof(CustomAttribute), true)?.Any() ?? false)
        return true;
}
PTD
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lisandro101
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2

Try

if (context.Filters.Any(x => x.GetType() == typeof(Microsoft.AspNetCore.Mvc.Authorization.AllowAnonymousFilter)))
            return;
Delmirio Segura
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