Exception in runnning application

INFO: validateJarFile(C:\Program Files\Apache Software Foundation\Tomcat 6.0\webapps\birt\WEB-INF\lib\servlet-api.jar) - jar not loaded. See Servlet Spec 2.3, section 9.7.2. Offending class: javax/servlet/Servlet.class

How can we solve this issue?Actually what is the reason for this?Can u please help me. I'm Stuck up here.

Thanks in advance,

View Answers

February 21, 2011 at 11:59 AM

Have you put the servlet-api jat in Tomcat 6.0/lib? Anyways,

Follow these steps:

Put servlet-api.jar inside the lib folder of apache tomcat. 1)create a servlet.

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;

public class HelloWorld extends HttpServlet{ 
  public void doGet(HttpServletRequest request, HttpServletResponse response)
                                   throws ServletException,IOException{
    response.setContentType("text/html");
    PrintWriter pw = response.getWriter();
    pw.println("<html>");
    pw.println("<head><title>Hello World</title></title>");
    pw.println("<body>");
    pw.println("<h1>Hello World</h1>");
    pw.println("</body></html>");
  }
}

2)Go to the webapps folder of your apache tomcat and create a web application folder but it should having an appropriate name like examples.

3)Create web.xml and classes folder inside the WEB_INF folder of web application folder.

4)Copy the servlet to the classes folder.

5)Edit the web.xml to include servlet?s name and url pattern.

 <servlet>
  <servlet-name>HelloWorld</servlet-name>
  <servlet-class>HelloWorld</servlet-class>
 </servlet>
 <servlet-mapping>
 <servlet-name>HelloWorld</servlet-name>
 <url-pattern>/HelloWorld</url-pattern>
 </servlet-mapping>

6)Compile your servlet.

7)Run Tomcat server by clicking the startup.bat file. This is located inside the bin folder of apache tomcat.

8)Open the browser and type the following url:

http://localhost:8080/webapplicationfolder_name/HelloWorld

For more information, visit the following link:

Servlet Tuorials

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