Actually, the code I was looking for was the part where I get the file that is passed to the servlet... The query string would look something like this:
http://localhost:8080/servletname?pg=/pages/page.html Where the server root may be c:\www\myweb So I would want the pg variable to be equal to: c:\www\myweb\pages\page.html I did accomplish this with this line of code: out.println(readTemplate(getServletContext().getRealPath("/" + pg))); The ReadTemplate function actually reads in the file and passes it back... If anyone has more efficient code For the following, please share: public String readTemplate(String pg) { File f = new File(pg); int l = (int)f.length( ); byte[ ] buffer = new byte[l]; try { FileInputStream in = new FileInputStream(pg); in.read(buffer, 0, l); } catch(Exception e) { System.out.println("Exception: " + e.getMessage( ) + "has occurred"); } String s = new String(buffer); return s; } -Wil Moore -----Original Message----- From: A mailing list for discussion about Sun Microsystem's Java Servlet API Technology. [mailto:[EMAIL PROTECTED]] On Behalf Of RBonazzo Sent: Friday, March 08, 2002 12:55 AM To: [EMAIL PROTECTED] Subject: R: Importance: High You can use this for example java.io.BufferedReader in = new java.io.BufferedReader(new java.io.FileReader("c:\\odd_directory\\web\\templates\\page.html")); do { Riga = in.readLine(); if (Riga == null) { in.close(); break; } ContenutoFile = ContenutoFile + Riga + "\n"; } while (true); Hope that it can be help you. Regards Rinaldo [EMAIL PROTECTED] -----Messaggio originale----- Da: A mailing list for discussion about Sun Microsystem's Java Servlet API Technology. [mailto:[EMAIL PROTECTED]] Per conto di wilbert moore Inviato: gioved� 7 marzo 2002 21.22 A: [EMAIL PROTECTED] Oggetto: What I'm trying to do is probably pretty simple. Its pretty easy to do in VB/ASP so I know this will be a breeze for someone to answer. My servlet is to read in an html file's contents which is sitting in a folder under the main server's root. Since I have not yet been successful in configuring tomcat to work under apache's control, my servlet is at http://localhost:8080/servletname If my html file is at http://localhost:80/page.html, what is the call I need to make to find the raw directory (eg. c:\odd_directory\web\templates\ ) I could obviously code this in hard, which I did to test so far, however, if the app were moved, this would not work. Thanks for the help, and I hope I supplied enough info for this simple task. __________________________________________ Launch your own web site Today! Create a Web site for your family, friends, photos, or a special event. Visit: http://www.namezero.com/sitebuilder ________________________________________________________________________ ___ To unsubscribe, send email to [EMAIL PROTECTED] and include in the body of the message "signoff SERVLET-INTEREST". Archives: http://archives.java.sun.com/archives/servlet-interest.html Resources: http://java.sun.com/products/servlet/external-resources.html LISTSERV Help: http://www.lsoft.com/manuals/user/user.html ________________________________________________________________________ ___ To unsubscribe, send email to [EMAIL PROTECTED] and include in the body of the message "signoff SERVLET-INTEREST". Archives: http://archives.java.sun.com/archives/servlet-interest.html Resources: http://java.sun.com/products/servlet/external-resources.html LISTSERV Help: http://www.lsoft.com/manuals/user/user.html ___________________________________________________________________________ To unsubscribe, send email to [EMAIL PROTECTED] and include in the body of the message "signoff SERVLET-INTEREST". Archives: http://archives.java.sun.com/archives/servlet-interest.html Resources: http://java.sun.com/products/servlet/external-resources.html LISTSERV Help: http://www.lsoft.com/manuals/user/user.html
