3_使用工具快速生成servlet

最新推荐文章于 2023-11-27 11:04:31 发布

原创最新推荐文章于 2023-11-27 11:04:31 发布 · 272 阅读

0 ·

CC 4.0 BY-SA版权

MVC 专栏收录该内容

7 篇文章

订阅专栏

本文介绍如何在Eclipse中快速创建Servlet，通过具体代码示例展示了doGet和doPost方法的使用，并提供了index.jsp页面实现GET和POST请求的调用。

Dmoe_01.java

package study;

import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

/**
 * *使用eclipse快速生成servlet,在新建时搜索servlet文件,填上包名类名和需要重写的一些方法
 * 删除一些相关注释后即可
 */
@WebServlet("/Demo_01")
public class Demo_01 extends HttpServlet {
	private static final long serialVersionUID = 1L;
	
	protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
		System.out.println("快速生成servlet的doget");
	}

	protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
		System.out.println("快速生成的sevlet的dopost");
	}

}

index.jsp

<%@ page language="java" contentType="text/html; charset=utf-8"
    pageEncoding="utf-8"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>Insert title here</title>
</head>
<body>
	<a href="Demo_01">doGet</a><!-- 超链接都是get方式 -->
	<form action="Demo_01" method="post">
		<input type="submit" value="dopost">
	</form>
</body>
</html>