Microsoft.Xna.Framework.TitleContainer.GetFilename C# (CSharp) Method

        internal static string GetFilename(string name)
        {
            return FileHelpers.NormalizeFilePathSeparators(new Uri("file:///" + name).LocalPath.Substring(1));
        }
    }

        public static Stream OpenStream(string name)
        {
            string filename = TitleContainer.GetFilename(name);

            if (Path.IsPathRooted(filename))
            {
                throw new ArgumentException("Invalid filename. TitleContainer.OpenStream requires a relative path.");
            }
            else
            {
                return((Stream)File.OpenRead(Path.Combine(TitleContainer.Location, filename)));
            }
        }