Hi all,
I'm trying to resolve this problem with all the solutions that you gave
me, but it doesn't work...
This is what I did:
in my java bean (not a servlet), I have this code:
public class DbBean {
public int Connect() {
InputStream is =
Thread.currentThread().getContextClassLoader().getResourceAsStream("config.txt");
if (is == null) {
return 0;
}
else {
return 1;
}
}
then in my jsp, I called this method, and then I write the value (0 or
1)..
The txt file is in "WEB-INF/classes/beans...", because "DbBean" is in a
package called "beans", and I start tomcat from TOMCAT_HOME/bin..
When I load the jsp, the method Connect of the DbBean (java bean) returned
0, which means the InputStream is null, but if I put the txt file in
TOMCAT_HOME/bin, I had no problem...., the method returned 1.... why is
that??.. I'm using Tomcat 3.2
Do I need to set something else in Tomcat??....
thanks again
Alex Tomita
"Drinkwater, GJ (Glen)" <[EMAIL PROTECTED]>
13/08/2002 08:11 a.m.
Please respond to Tomcat Users List
To: Tomcat Users List <[EMAIL PROTECTED]>
cc:
Subject: RE: Quick Question
tomcats default directory is where ever you called the startup.sh/bat
file.
So if you dont use another script to call the startup.sh/bat file tomcats
default will be the bin directory.
Use this code to find the directory where WEB-INF is. Then you can
traverse
your directory structure from there.
//get context path
ServletConfig scon = null ;
String workingDir = null;
public void init(ServletConfig config) {
scon = config ;
}
public void doPost(HttpServletRequest request, HttpServletResponse
response)
throws ServletException, IOException {
//get working dir
ServletContext sc = scon.getServletContext();
workingDir = sc.getRealPath("");
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