Thankyou Sachin. Thankyou Kavitha. Now html is opening with following:
http://localhost:8080//mySecondWebProject/WebContent/form.html ***Note: i am doing all from commond prompt.
but when i select submit query i got following error:
HTTP Status 404 - //mySecondWebProject/WebContent/SelectBeer.do
-----------------------------------------------------------------------
type Status report
message //mySecondWebProject/WebContent/SelectBeer.do
description The requested resource (//mySecondWebProject/WebContent/SelectBeer.do) is not available.
------------------------------------------------------------------------
Apache Tomcat/6.0.18
My Paths are:
C:\Tomcat 6.0\webapps\mySecondWebProject\src\BeerSElect.java
C:\Tomcat 6.0\webapps\mySecondWebProject\WebContent\form.html
C:\Tomcat 6.0\webapps\mySecondWebProject\WebContent\WEB-INF
C:\Tomcat 6.0\webapps\mySecondWebProject\WebContent\WEB-INF\classes\BeerSelect.class
C:\Tomcat 6.0\webapps\mySecondWebProject\WebContent\WEB-INF\web.xml
and my codes are:
SelectBeer.java
---------------
import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;
public class BeerSelect extends HttpServlet{
public void doPost(HttpServletRequest request,HttpServletResponse response)throws IOException,ServletException{
response.setContentType("text/html");
PrintWriter out=response.getWriter();
out.println("Beer Selection Advice
");
String c= request.getParameter("color");
out.println("
Got beer color"+c);
}
}
web.xml
--------
- <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/web-app_2_4.xsd" version="2.4">
- <servlet>
<servlet-name>Ch3 Beer</servlet-name>
<servlet-class>BeerSelect</servlet-class>
</servlet>
- <servlet-mapping>
<servlet-name>Ch3 Beer</servlet-name>
<url-pattern>/SelectBeer.do</url-pattern>
</servlet-mapping>
</web-app>
please help me further.
Thank you,