servlet web.xml配置的问题

Question

package servlet;import java.io.IOException;import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;public class TestServlet extends HttpServlet{
public TestServlet(){}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String username=request.getParameter("username");
UserManage usermanage=new UserManage();
usermanage.addUser(username);

request.getRequestDispatcher("/addSuccess1").forward(request, response);

}

@Override
protected void doPost(HttpServletRequest  request, HttpServletResponse response) throws ServletException, IOException {
doGet(request,response);
}}
配置：
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.4" 
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">
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
  <servlet-name>TestServlet</servlet-name>
  <servlet-class>servlet.TestServlet</servlet-class>
  
  
  </servlet>
  
  <servlet-mapping>
  <servlet-name>TestServlet</servlet-name>
  
  <url-pattern>/servlet/TestServlet</url-pattern>
  </servlet-mapping>
 
  
</web-app>结果显示找不到路径，这是怎么回事呢，配置哪错了

Answer 1

信息有点少，不好分析原因。
addSuccess1是个servlet吗？它在xml里的配置中，url-pattern应该是/addSuccess1.
  <servlet>
    <servlet-name>addSuccess1</servlet-name>
<servlet-class>...</servlet-class>  </servlet>
  <servlet-mapping>
    <servlet-name>addSuccess1</servlet-name>
    <url-pattern>/addSuccess1</url-pattern>
  </servlet-mapping>

Answer 2

不知道您要问什么？
貌似是请求路径的问题
表单提交action应该是servlet/TestServlet

Answer 3

request.getRequestDispatcher("/addSuccess1").forward(request, response);
----跳转是要交给JSP吧