（急，在线等）请教servlet问题

Question

我写了个servlet.java，编译后把它放在C:\Tomcat 5.0\webapps\shop\WEB-INF\classes\com\shop下，然后在jsp文件中写<form id="form1" name="form1" method="post" action="FormAction">，为什么运行后显示The requested resource (/shop/com/shop/FormAction) is not available.
，请高手指点一下，谢谢 
package com.shop;import java.io.IOException;
import java.util.Collection;import javax.servlet.ServletException;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;/**
 * Servlet implementation class for Servlet: UserListServlet
 */
public class FormAction extends javax.servlet.http.HttpServlet implements
javax.servlet.Servlet
{
/*
 * (non-Java-doc)
 *
 * @see javax.servlet.http.HttpServlet#HttpServlet()
 */
public FormAction()
{
super();
} /*
 * (non-Java-doc)
 *
 * @see javax.servlet.http.HttpServlet#doGet(HttpServletRequest request,
 *      HttpServletResponse response)
 */
protected void doGet(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException
{
response.setHeader("Pragma","No-cache");
response.setHeader("Cache-Control","no-cache");
response.setDateHeader("Expires", 0);
String name = request.getParameter("name");
int age = Integer.parseInt(request.getParameter("age"));
System.out.println ("name= "+name);
System.out.println ("age= "+age);
System.out.println("ok");     }
}

Answer 1

form id="form1" name="form1" method="post" action="FormAction">，
把post改成get看看

Answer 2

你得在web.xml里配置servlet啊
<web-app>
  <servlet>
    <servlet-name>FormAction</servlet-name>
    <servlet-class>com.shop.FormAction</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>FormAction</servlet-name>
    <url-pattern>/FormAction</url-pattern>
  </servlet-mapping>
</web-app>

Answer 3

还有  把你的formaction改成你的servlet类名!!

Answer 4

alieneagle() 正解，顶！！！