ServletContext的请求和转发

ServletContext的请求和转发

1、 新建class

 

 

 

 写类：

 1 package com.wang.servlet;
 2 
 3 import javax.servlet.RequestDispatcher;
 4 import javax.servlet.ServletContext;
 5 import javax.servlet.ServletException;
 6 import javax.servlet.http.HttpServlet;
 7 import javax.servlet.http.HttpServletRequest;
 8 import javax.servlet.http.HttpServletResponse;
 9 import java.io.IOException;
10 
11 public class ServletDemo04 extends HttpServlet {
12     @Override
13     protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
14         ServletContext context = this.getServletContext();
15         System.out.println("进入了ServletDemo04");
16         RequestDispatcher requestDispatcher = context.getRequestDispatcher("/gp");
17         requestDispatcher.forward(req,resp);
18     }
19 
20     @Override
21     protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
22         doGet(req, resp);
23     }
24 }
View Code

2、 注册和映射

 

 

1     <!--注册和映射-->
2     <servlet>
3         <servlet-name>sd4</servlet-name>
4         <servlet-class>com.wang.servlet.ServletDemo04</servlet-class>
5     </servlet>
6     <servlet-mapping>
7         <servlet-name>sd4</servlet-name>
8         <url-pattern>/sd4</url-pattern>
9     </servlet-mapping>
View Code

3、 修改Tomcat和运行

 

posted @ 2020-03-24 22:51  WZ_BeiHang  阅读(226)  评论(0)    收藏  举报