|
Hi All
I have written a Simple servlet
program its compilation is Ok .
my servlet server seems to work very nice ,
it tried successfully the examples servlets that come with
.
But each time i try to run my own it
results in a "not found page " message.
I have edited the property file to
set up my Servlet property name here is its content
:
# $Id: servlets.properties,v 1.2 1999/04/02
02:04:22 duncan Exp $
# Define servlets here
#
<servletname>.code=<servletclass>
# <servletname>.initparams=<name=value>,<name=value> snoop.code=SnoopServlet
snoop.initparams=initarg1=foo,initarg2=bar jsp.code=com.sun.jsp.runtime.JspServlet SimpleServlet.code=SimpleServlet ~
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ "servlets.properties" 12 lines, 322 characters Following is the starting of the server
:
[root@bcm600 jsdk2.1]# ./startserver&
[1] 15192 [root@bcm600 jsdk2.1]# Using classpath: ./server.jar:./servlet.jar:.:/usr/java/l ib/class.zip:/informix/jdbc/lib/ifxjdbc.jar:/informix/jdbc/lib/ifxjdbc-g.jar:/ho me/sidaty/servlet/jsdk2.1/servlet.jar:/home/sidaty/servlet/jsdk2.1/server.jar JSDK WebServer Version 2.1 Loaded configuration from file:/home/sidaty/servlet/jsdk2.1/default.cfg endpoint created: bcm600/172.16.4.60:8081 [1]+
Done
./startserver
[root@bcm600 jsdk2.1]# Can Anybody tell me what is wrong in my
property file , what should be its location (now it is
under
<install-dir>webpages/WEB-INF/servlets.properties
What should be the exact value in
sevlets.properties and default.cfg files and their path
.
A lot of thanks in advance
|
