Hi Louis,
First thing is that your class structure should preferable look like
WEB-INF/classes/com/iss/npd/servlets instead of
WEB-INF/com/iss/npd/servlets. Notice that the "classes" is missing in your
examples. Secondly you can get your current path in the servlet using
String appPath = getServletContext().getRealPath("/");
This will give you an OS path of your application, so if your properties
file is "Config.props" in the WEB-INF/classes/com/iss/npd directory then
just load the properties using
String configPath = appPath +
"/WEB-INF/classes/com/iss/npd/Config.props" ;
Hope this helps
Ajay Mahajan
Wipro Technologies
----- Original Message -----
Date: Mon, 18 Sep 2000 09:13:09 +0100
From: "Louis" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Subject: how to get the current path?
Message-ID: <004501c02148$4835c820$[EMAIL PROTECTED]>
Hi,
I got a servlet and it will call another class. In that class I need =
to read a properties file. But i don't know where to put my properties =
file, I try to put anywhere, but still got file not found error. So =
what method i can use to know the current path of that
class will reading?
for example,
The servlet testDB is in directory
WEB-INF /com/iss/npd/servlets
And the class file dbConnect is in directory
WEB-INF/com/iss/npd/database
The dbConnect will read the database.properties file, anyone can tell =
me where I can put this file?
In JMC, Servlet Definitions I give this Name=3DtestDB, Class Name =3D =
com.iss.npd.servlets.testDB
then in Servlet URL Mappings, I give the Virtual Path/Extension =3D =
/testDB, Servlet Invoked =3D testDB
Thanks
Louis
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