BetterCms.Module.Blog.Services.DefaultBlogMLService.DeserializeXMLFile C# (CSharp) Метод

        public BlogMLBlog DeserializeXMLFile(string filePath)
        {
            Stream fileStream;
            try
            {
                fileStream = File.OpenRead(filePath);
            }
            catch (Exception exc)
            {
                var message = "Failed to open file for reading";
                var logMessage = string.Format("Failed to open XML file {0} for reading.", filePath);
                throw new ValidationException(() => message, logMessage, exc);
            }

            return DeserializeXMLStream(fileStream);
        }