Rami Yusf
Rami Yusf

Reputation: 3032

Access Attribute data from method

I'm using asp.net core api project and I use a custom ActionFilter attribute to do some authentication validations as below:

public class LoggedInAttribute : ActionFilterAttribute
{
    public Login LoggedInUser { get; private set; }
    public override void OnActionExecuting(ActionExecutingContext con)
    {
        LoggedInUser = //here i get the logged in user(from a http token request header) and load it from database
        if (LoggedInUser == null)
        {
            con.Result = new UnauthorizedResult();
        }
    }
}

Then I placed this attribute on an action in the api controller as below:

    [HttpGet("user/GetAccountInfo")]
    [LoggedIn]
    public AccountInfoDTO GetAccountInfo()
    {
       //Here i want to get the placed [LoggedIn] instance to get it's LoggedInUser value
    }

I need to get the LoggedInUser property inside the method, I've tried some reflection but I get null everytime.

Upvotes: 0

Views: 193

Answers (1)

Brando Zhang
Brando Zhang

Reputation: 28257

According to your description, we couldn't directly read the LoggedInAttribute property in the controller action, they are different class.

If you want to get the login model, I suggest you could put it in httpcontext item and read the httpcontext item in the action.

More details, you could refer to below codes:

    public void OnActionExecuting(ActionExecutingContext context)
    {
        LoggedInUser = new Login { Id = 1 };

        context.HttpContext.Items["Login"] = LoggedInUser;
        
        //throw new NotImplementedException();
    }

Action:

  [HttpGet]
    public IEnumerable<WeatherForecast> Get()
    {
        var loginuser = HttpContext.Items["Login"];
        var rng = new Random();
        return Enumerable.Range(1, 5).Select(index => new WeatherForecast
        {
            Date = DateTime.Now.AddDays(index),
            TemperatureC = rng.Next(-20, 55),
            Summary = Summaries[rng.Next(Summaries.Length)]
        })
        .ToArray();
    }

Result:

enter image description here

Upvotes: 1

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