Mega Code Archive

 
Categories / Java Tutorial / Servlet
 

Servlet Controller Dispatcher

import java.io.*; import javax.servlet.*; import javax.servlet.http.*; import java.util.*; public class MyServlet extends HttpServlet {     public void doGet(HttpServletRequest request,          HttpServletResponse response)           throws ServletException, java.io.IOException {         RequestDispatcher dispatcher = null;         String param = request.getParameter("go");              if (param == null)                  throw new                       ServletException("Missing parameter in Controller.");              else if (param.equals("weather"))                  dispatcher = getServletContext().                      getNamedDispatcher("Weather");              else if (param.equals("maps"))                  dispatcher = getServletContext().                      getNamedDispatcher("Maps");             else                 throw new ServletException(                     "Improper parameter passed to Controller.");          /*check for a null dispatcher, then              dispatch the request to the correct URL*/         if (dispatcher != null)             dispatcher.forward(request,response);         else             throw new ServletException(               "Controller received a null dispatcher.");     } } <?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>MyServletName</servlet-name>              <servlet-class>MyServlet</servlet-class>         <init-param>             <param-name>                 go             </param-name>             <param-value>                 weather             </param-value>         </init-param>                   </servlet>          <servlet-mapping><servlet-name>MyServletName</servlet-name>         <url-pattern>/index.html</url-pattern>     </servlet-mapping> </web-app>