Need help with web.xml file

[northfield Sid]

I want to create an web.xml page
I frequently access a servlet page with this command:
response.sendRedirect("http://localhost:8080/myJSPs/servlet/portalftp.ServletFaillogin");
Is it possible to avoid typing this tedious line by using the web.xml file?
Here is what I have tried:

<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
    PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.2//EN"
    "http://java.sun.com/j2ee/dtds/web-app_2_2.dtd">
<web-app>
<servlet>
<servlet-name>faillog</servlet-name>
<servlet-class>portalftp.ServletFaillogin</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>faillog</servlet-name>
<url-pattern>/faillog</url-pattern>
</servlet-mapping>
</web-app>

so thus the code now becomes:
response.sendRedirect("http://localhost:8080/myJSPs/servlet/faillog");
2nd question:
I have put the web.xml file at this location:
C:\tomcat\jakarta-tomcat-3.3.1\webapps\myJSPs-XML\WEB-INF
The file ServletFaillogin.java and rest of Java pages are at:
C:\tomcat\jakarta-tomcat-3.3.1\webapps\myJSPs-XML\WEB-INF\classes\portalftp
and all my HTML and JSP pages are at:
C:\tomcat\jakarta-tomcat-3.3.1\webapps\myJSPs-XML\jsp\portal-project

[Matthew Phillips]

With the servlet-mapping tag in place, you shouldn't need /servlet in your url. It also appears that you have a problem with your context. Your webapp folder is called myJSPs-XML, but you are typing myJSPs in your URL. Unless you have changed your context in the Tomcat configuration file, that shouldn't work.