This week's book giveaway is in the OCAJP 8 forum. We're giving away four copies of OCA Java SE 8 Programmer I Study Guide and have Edward Finegan & Robert Liguori on-line! See this thread for details.
Hi, I have a simple servlets/jsp application running under tomcat 4.3.1 in linux environment. Currently all my jsp's are located under jsp directory under application root.
To access my home jsp, for my test context root of the application, I have to write URL; ...localhost:8080/test/jsp/home.jsp I would like to use hide my directory structure and the jsp file names and call the URL as: ...localhost:8080/test/home which should get me the same jsp page. I have tried several ways and tried to make the following change in my web.xml but no luck. <servlet> <servlet-name>homeJSP</servlet-name> <display-name>home</display-name> <jsp-file>/jsp/home.jsp</jsp-file> </servlet> <servlet-mapping> <servlet-name>homeJSP</servlet-name> <url-pattern>/home</url-pattern> </servlet-mapping>
Using the URL: ...localhost:8080/test/home
give me the error: org.apache.jasper.JasperException at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:207) ...
However, if I use the URL ...localhost:8080/test/jsp/home.jsp It works well. Your help will be appreciated. Thanks.
[ March 24, 2006: Message edited by: B Bhutta ] [ March 24, 2006: Message edited by: B Bhutta ]