System.Xml.Xsl.SourceLineInfo.GetFileName C# (CSharp) Method

        public static string GetFileName(string uriString) {
            Uri uri;

            if (uriString.Length != 0 &&
                System.Uri.TryCreate(uriString, UriKind.Absolute, out uri) &&
                uri.IsFile
            ) {
                return uri.LocalPath;
            }
            return uriString;
        }
    }

 private static string AppendLineInfoMessage(string message, ISourceLineInfo lineInfo)
 {
     if (lineInfo != null)
     {
         string fileName        = SourceLineInfo.GetFileName(lineInfo.Uri);
         string lineInfoMessage = CreateMessage(SR.Xml_ErrorFilePosition, fileName, lineInfo.Start.Line.ToString(CultureInfo.InvariantCulture), lineInfo.Start.Pos.ToString(CultureInfo.InvariantCulture));
         if (lineInfoMessage != null && lineInfoMessage.Length > 0)
         {
             if (message.Length > 0 && !XmlCharType.Instance.IsWhiteSpace(message[message.Length - 1]))
             {
                 message += " ";
             }
             message += lineInfoMessage;
         }
     }
     return(message);
 }