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Home/ Questions/Q 6014089
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T02:38:30+00:00 2026-05-23T02:38:30+00:00

I have the follow method that returns a dynamic object representing an IEnumerable<‘a> (‘a=anonymous

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I have the follow method that returns a dynamic object representing an IEnumerable<'a> (‘a=anonymous type) :

    public dynamic GetReportFilesbyStoreProductID(int StoreProductID)
    {
        Report report = this.repository.GetReportByStoreProductID(StoreProductID);

        if (report == null || report.ReportFiles == null)
        {
            return null;
        }

        var query = from x in report.ReportFiles
                    orderby x.DisplayOrder
                    select new { ID = x.RptFileID, Description = x.LinkDescription, File = x.LinkPath, GroupDescription = x.ReportFileGroup.Description };

        return query;
    }

I want to be able to access the Count property of this IEnumerable anonymous type. I’m trying to access the above method using the following code and it is failing:

        dynamic Segments = Top20Controller.GetReportFilesbyStoreProductID(StoreProductID");

        if (Segments.Count == 0)  // <== Fails because object doesn't contain count.
        {
            ...
        }
  • How does dynamic keyword operate?
  • How can I access the Count property of the IEnumerable anonymous type?
  • Is there a way I can use this anonymous type or do I have to create a custom object so that I can pass back a strongly-typed IEnumerable<myObject> instead of dynamic?

I’d prefer to not do that if I can as this method is only called in one place and creating a throw-away object seems like overkill.

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  1. Editorial Team
    Editorial Team
    2026-05-23T02:38:31+00:00Added an answer on May 23, 2026 at 2:38 am

    The IEnumerable<T> that’s returned from that method doesn’t have a Count property, so I don’t know what you’re talking about. Maybe you forgot to write ToList() on the end to reify it into a list, or you meant to call the Count() method on IEnumerable<T>?

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