I've been struggling with this web xml file for Tomcat for a while.
<context-param>
<param-name>name</param-name>
<param-value>Bob</param-value>
</context-param>
<servlet>
<servlet-name>test</servlet-name>
<servlet-class>TestServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>test</servlet-name>
<url-pattern>/servlet/*</url-pattern>
</servlet-mapping>
I know that this file is being read because when i test using
public class TestServlet extends HttpServlet{
public void doGet(HttpServletRequest req, HttpServletResponse res){
res.setContentType("text/html");
String name = getServletContext().getInitParameter("name");
PrintWriter out = null;
try{
out = res.getWriter();
}catch(Exception e){}
out.println("<html><head></head><body><img src=\"/twitter.png\"><p>Hi my name is " + name + "</p></body></html>");
}
}
I am able to read the name I put into the context-param. My question for you guys is how do I create a URL mapping such that I do not have to go through /servlet/ to access my servlets in the URL? When I try to make a url-pattern such as
/test/*, i cannot access the servlet even if i say website/test/TestServlet. I get a 404 Not Found error from the browser
