Page cannot be dispalyed message

Question

i installed tomcat and created a simple servlet named servlet1.
i created a directory named servlet1 and put servlet1.class under servlet1/WEB-INF/classes. i also put web.xml under WEB-INF.
in web.xml i wrote-
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE web-app
PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
"http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<servlet>
<servlet-name>servlet1</servlet-name>
<display-name>servlet1</display-name>
<servlet-class>servlet1</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>servlet1</servlet-name>
<url-pattern>/servlet/servlet1</url-pattern>
</servlet-mapping>
</web-app>
but when i tried to run it in the browser with url's like-
http://localhost:8080/servlet/servlet1
i got the "the page cannot be displayed" message
(i also tried without the 8080).
my environment variables are-
JAVA_HOME = C:\j2sdk1.4.2_10\bin
PATH = %PATH%;%JAVA_HOME%\bin
what is wrong?
thanks in advanced.

Answer 1

hi friend,
gave the url in browser like this :
http://localhost:8080/servlet1/servlet/servlet1
now ur application will get executed. Otherwise I gave small example with the required programs. pls look at those once and make changes as per my instructions. if u have any doubt let me know immediately.
make the application as follows: u can execute ur application:
i will give u each and everything clearly, follow these.
step 1:[b]write the servlet : For example : TestServlet
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.GenericServlet;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
* @author sivakumar
public class TestServlet extends GenericServlet  {
      public void service(ServletRequest request,ServletResponse response)throws ServletException,IOException{
               PrintWriter out=response.getWriter();
               response.setContentType("text/html");
               out.println("Hi this is first servlet");
step 2:[b]web.xml
<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"
    version="2.4">
  <servlet>
    <servlet-name>testServlet</servlet-name>
    <servlet-class>TestServlet</servlet-class>
  </servlet>
  <!-- Standard Action Servlet Mapping -->
  <servlet-mapping>
    <servlet-name>testServlet</servlet-name>
    <url-pattern>/testServlet</url-pattern>
  </servlet-mapping>
</web-app>
Step 3 : set classpath to servlet.jar or servlet-api.jar depending upon ur tomcat version
for example, if Tomcat5.0 is available in C:\ drive,then
set CLASSPATH=C:\Tomcat5.0\common\lib\servlet-api.jar
Step 4: Compile your Servlet
Step 5: Create your directory structure as follows : and place ur class file in classes directory of WEB-INF
          RootDirecotry
                         WEB-INF
                                     classes
                                     lib
                                     web.xml
                           html [if needed]
                           jsp [if needed]
        For ur application( For example ur root direcotory is   servlet)                 
             servlet
                         WEB-INF
                                     classes
                                                TestServlet.class
                                      lib
                                      web.xml  (which i have given above)      
Step 6 : Deploy this application in tomcat's webapps directory
Step 7 : Give the URL as follows
http://localhost:8080/servlet/testServlet
Port number depends on configuration. Make sure ur tomcat is running at 8080. otherwise give the port number which u mentioned at the time of tomcat installation.
Thanks ,
Regards.
Message was edited by:
siva_forums