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servlet入门(1)

2015-10-20 09:43 141 查看
我使用的是eclipse+tomcat+mysql的开发环境

1,新建一web项目,过程中切勿忘记点选创建web.xml文件

2,在java Resources中创建servlet

3,在webcontent文件夹中创建一个jsp页面,下面是我的第一个jsp页面的信息

<%@ page language="java" contentType="text/html; charset=uf-8"
pageEncoding="ISO-8859-1"%>
<!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=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<h1>this is my first servlet programe</h1>
<hr>
<a href="servlet/HelloServlet">处理get请求</a>
</body>
</html>


4,之后开始写sevlet的程序(重写doget()方法)

package servlet;

import java.io.IOException;
import java.io.PrintWriter;

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

/**
* Servlet implementation class HelloServlet
*/
@WebServlet("/HelloServlet")
public class HelloServlet extends HttpServlet {
private static final long serialVersionUID = 1L;

/**
* @see HttpServlet#HttpServlet()
*/
public HelloServlet() {
super();
// TODO Auto-generated constructor stub
}

/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
response.getWriter().append("Served at: ").append(request.getContextPath());
System.out.println("处理get请求");
PrintWriter out = response.getWriter();
out.print("<strong>hello servlet</strong><br>");
}

/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
doGet(request, response);
}

}


5,在web.xml中配置sevlet信息

增加和信息

之后运行就可以了
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