Click here to Skip to main content
Licence CPOL
First Posted 17 Sep 2003
Views 72,295
Bookmarked 29 times

XML Directory Tree Generator

By | 17 Sep 2003 | Article
Describes a directory tree documenter that outputs XML.

Introduction

This code snippet recursively enumerates all the folders and files under a given starting folder and generates an XML document as an output, that represents the same hierarchy as of the traversed directories. The resulting XML document can then be transformed with a simple XSLT transformation to generate the desired output. The main reason for writing this was to display the contents of a given directory on a website, without having to enable directory browsing on the server. I wanted to be able to define certain filter criteria to control the output generated, so it was important to capture as much information pertaining to the folders and files as possible.

Using the code

The code is implemented as FileSystemInfoLister class. The main entry point is the GetFileSystemInfoList() method that returns an XML document. The AddElements() method is called recursively until the entire path is traversed.

using System;
using System.Xml;
using System.IO;


namespace Hosca.FileSystemInfoLister {

    public class FileSystemInfoLister {

        XmlDocument xmlDoc;

        public FileSystemInfoLister() {
            xmlDoc = new XmlDocument();
        }

        public XmlDocument GetFileSystemInfoList(string StartFolder) {

            try{
                XmlDeclaration xmlDec = xmlDoc.CreateXmlDeclaration("1.0", 
                                                                "", "yes");
                xmlDoc.PrependChild ( xmlDec );
                XmlElement nodeElem =  XmlElement("folder",
                         new DirectoryInfo(StartFolder).Name); 
                xmlDoc.AppendChild(AddElements(nodeElem,StartFolder));
            }
            catch(Exception ex) {
                xmlDoc.AppendChild(XmlElement("error",ex.Message));
                return xmlDoc;
            }
            return xmlDoc;
        }


        private XmlElement AddElements(XmlElement startNode, 
                                               string Folder){
            try{
                DirectoryInfo dir = new DirectoryInfo(Folder);
                DirectoryInfo[] subDirs = dir.GetDirectories() ;
                FileInfo[] files = dir.GetFiles();
                foreach(FileInfo fi in files){
                  XmlElement fileElem = XmlElement("file",fi.Name);
                  fileElem.Attributes.Append(XmlAttribute("Extension", 
                                                           fi.Extension));
                  fileElem.Attributes.Append(XmlAttribute("Hidden", 
                   ((fi.Attributes & FileAttributes.Hidden) != 0) 
                                                       ? "Y":"N"));
                  fileElem.Attributes.Append(XmlAttribute("Archive", 
                   ((fi.Attributes & FileAttributes.Archive ) != 0) 
                                                       ? "Y":"N"));
                  fileElem.Attributes.Append(XmlAttribute("System", 
                   ((fi.Attributes & FileAttributes.System ) != 0) 
                                                       ? "Y":"N"));
                  fileElem.Attributes.Append(XmlAttribute("ReadOnly", 
                   ((fi.Attributes & FileAttributes.ReadOnly ) != 0) 
                                                        ? "Y":"N"));
                  startNode.AppendChild(fileElem);
                }
                foreach (DirectoryInfo sd in subDirs) {
                  XmlElement folderElem = XmlElement("folder",sd.Name);
                  folderElem.Attributes.Append(XmlAttribute("Hidden", 
                    ((sd.Attributes & FileAttributes.Hidden) != 0) 
                                                       ? "Y":"N"));
                  folderElem.Attributes.Append(XmlAttribute("System", 
                    ((sd.Attributes & FileAttributes.System  ) != 0) 
                                                       ? "Y":"N"));
                  folderElem.Attributes.Append(XmlAttribute("ReadOnly",
                    ((sd.Attributes & FileAttributes.ReadOnly ) != 0)
                                                       ? "Y":"N"));
                  startNode.AppendChild(AddElements(folderElem,
                                                     sd.FullName));
                }
                return startNode;
            }
            catch(Exception ex) {
                return XmlElement("error",ex.Message);
            }
        }
        private XmlAttribute XmlAttribute(string attributeName, 
                                          string attributeValue){
            XmlAttribute xmlAttrib = 
                xmlDoc.CreateAttribute(attributeName);
            xmlAttrib.Value = FilterXMLString(attributeValue);
            return xmlAttrib;
        }
        private XmlElement XmlElement(string elementName, 
                                       string elementValue){
            XmlElement xmlElement = xmlDoc.CreateElement(elementName);
            xmlElement.Attributes.Append(XmlAttribute("name", 
                                      FilterXMLString(elementValue)));
            return xmlElement;
        }
        private string FilterXMLString(string inputString){
            string returnString = inputString;
            if (inputString.IndexOf("&") > 0){
                returnString = inputString.Replace("&","&");
            }
            if (inputString.IndexOf("'") > 0){
                returnString = inputString.Replace("'","'");
            }
            return returnString;
        }    
    }
}

Points of interest

The only minor point of interest is that there are 2 characters that are allowable in file names but not in XML documents. In fact there are 5 characters that are not allowed in XML documents. They are :

< < less than
> > greater than
& & ampersand
&apos; ' apostrophe
" " quotation mark

Fortunately for us, Windows takes care of 3 of them. However the ampersand and the apostrophe which are legal to use in file names will have to be handled manually. The FilterXMLString method takes care of the details.

History

  • Sept 18 2003 - Initial version (1.0.0)

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

About the Author

Erhan Hosca

Web Developer

United States United States

Member



Sign Up to vote   Poor Excellent
Add a reason or comment to your vote: x
Votes of 3 or less require a comment

Comments and Discussions

 
You must Sign In to use this message board. (secure sign-in)
 
Search this forum  
 FAQ
    Noise  Layout  Per page   
  Refresh
GeneralMy vote of 1 Pinmembermovhlps5:20 29 Nov '08  
QuestionEditing XML file PinmemberKrishnakumar Bhagwat21:17 21 Aug '06  
HI Ppl,
 
I have an XML file as below..
XmlDocument xmlDoc = new XmlDocument();
XmlNode xn = xmlDoc.CreateElement("Recent");
xmlDoc.AppendChild(xn);
XmlDeclaration xd = xmlDoc.CreateXmlDeclaration("1.0", "utf-8", null);
XmlNode root = xmlDoc.SelectSingleNode("/Recent");
XmlElement mainNode = xmlDoc.CreateElement("Query");
XmlAttribute Name_attr = xmlDoc.CreateAttribute("Name");
XmlAttribute Scr_attr = xmlDoc.CreateAttribute("Script");
 
i = 1;
Name_attr.InnerText = recent_query+(i);
i = i + 1;
Scr_attr.InnerText = cmdText;
 
mainNode.Attributes.Append(Name_attr);
mainNode.Attributes.Append(Scr_attr);
root.AppendChild(mainNode);
xmlDoc.Save("C:\\xml\\recent.xml");
I am using ASP.NET...
Problem I am facing is that.. whenevr I run this program, old value is replaced by the new value I am giving.. but I would like to retain the old value present in that and add the new value to it and store the same. Please let me know how do I do that...
My text value which is to be stored will be present in "cmdText".
 
Krishna

Answerapp PinmemberMember 45204221:27 6 Dec '09  
GeneralSmall bug in FilterXMLString PinmemberRichard Poole6:15 12 May '04  
GeneralRe: Small bug in FilterXMLString PinmemberErhan Hosca7:15 12 May '04  
GeneralRe: Small bug in FilterXMLString Pinmemberdjarn22:34 29 Oct '04  
GeneralHandling Ampersands PinmemberDonald Xie15:11 11 Jan '04  
GeneralRe: Handling Ampersands PinmemberErhan Hosca3:29 12 Jan '04  
GeneralVery cool - I want more! PinsussShem Stimpy3:32 25 Sep '03  

General General    News News    Suggestion Suggestion    Question Question    Bug Bug    Answer Answer    Joke Joke    Rant Rant    Admin Admin   

Use Ctrl+Left/Right to switch messages, Ctrl+Up/Down to switch threads, Ctrl+Shift+Left/Right to switch pages.

Permalink | Advertise | Privacy | Mobile
Web03 | 2.5.120529.1 | Last Updated 18 Sep 2003
Article Copyright 2003 by Erhan Hosca
Everything else Copyright © CodeProject, 1999-2012
Terms of Use
Layout: fixed | fluid