Listing folders on a server

Question

Hello,
I have been scrounging the forums for any ideas on how I can list the folders on my server from my clients location.
I am making a user manual that's just a set of videos embedded into HTML files and I thought if I put each HTML file into sub folders then I could build a tree structure from the clients location and they could choose which topic they wish to view.
The reason for this is that the program is huge and making a user manual will take me months so I thought I'd make it gradually and also have the freedom to replace videos when the system changes.
The program that reads the folders and launches the HTML browser works fine on my machine but alas I didn't realise that you can't just use file.list() and get the folders from my server.
Is there some other way it can be done simply, I guess I could have an XML page of the folder structure in a fixed location that I could update when I put a new folder on the server but my boss would be extra pleased if I could do away with it and just whack the folders up when they are made.
Thanks
kind regards, Aaron :)

Answer 1

Simple html directory scraper using TagSoup:import  java.io.*;
import  java.net.*;
import  java.util.*;
// TagSoup HTML SAX parser, see http://mercury.ccil.org/~cowan/XML/tagsoup/
import  org.ccil.cowan.tagsoup.Parser;
import  org.xml.sax.Attributes;
import  org.xml.sax.InputSource;
import  org.xml.sax.SAXException;
import  org.xml.sax.helpers.DefaultHandler;
* Gets an http url, scans for links that may under this one.
public class HttpDir {
   * Maybe lists the child nodes in a folder, if it's the sort of html page
   * most web servers throw up for a directory listing.
   * @param url The URL of the page to scrape folder data from.
   * @return A list of possible sublinks.
   * @throws SAXException if something happens in the parsing.
   * @throws IOException if something happens in the streaming.
   * @throws IllegalArgumentException if url doesn't start with "http://"
  public static List listDir (final String url) throws SAXException, IOException {
    final List sublinks = new ArrayList();
    if (!url.startsWith("http://")) {
      throw new IllegalArgumentException("No an http URL '" + url + "'");
    final String dirUrl = url.endsWith("/") ? url : url + "/";
    final String serverbase = dirUrl.substring(0, dirUrl.indexOf("/", 7));
    final String dirPath = dirUrl.substring(serverbase.length());
    InputStream in = null;
    try {
      in = new URL(dirUrl).openStream();
      final Parser parser = new Parser();
      parser.setContentHandler(new DefaultHandler() {
        public void startElement (String namespaceURI, String localName, String qName, Attributes attrs) {
          if ("a".equals(localName)) {
            final String href = attrs.getValue("href");
            if (href != null) {
              if (href.startsWith("/")) {
                if (href.startsWith(dirPath)) {
                  // relative url, begining in '/', stemming from directory base
                  sublinks.add(href.substring(dirPath.length()));
              } else if (href.startsWith(dirUrl) && (href.length() > dirUrl.length())) {
                // full url, starting from directory base, but not the directory base
                sublinks.add(href.substring(dirUrl.length()));
              } else if (href.indexOf(':') == -1) {
                // not begining in '/' and no colon => no protocol, therefore relative url,
                // so is automatically under directory base
                sublinks.add(href);
      parser.parse(new InputSource(in));
    } finally {
      if (in != null) { in.close(); }
    return sublinks;
}