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Categories / Java Tutorial / Servlet

 

Forward to another location

import java.io.*; import javax.servlet.*; import javax.servlet.http.*; import java.util.*; public class MyServlet extends HttpServlet {   static final String NEW_HOST = "http://www.rntsoft.com";   public void doGet(HttpServletRequest req, HttpServletResponse res)                                throws ServletException, IOException {     res.setContentType("text/html");     PrintWriter out = res.getWriter();     String newLocation = NEW_HOST;     res.setHeader("Refresh", "10; URL=" + newLocation);     out.println("The requested URI has been moved to a different host.<BR>");     out.println("Its new location is " + newLocation + "<BR>");     out.println("Your browser will take you there in 10 seconds.");   } } <?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>                   </servlet>          <servlet-mapping><servlet-name>MyServletName</servlet-name>         <url-pattern>/index.html</url-pattern>     </servlet-mapping> </web-app>