【servlet学习1】使用eclipse+tomcat开发servlet示例

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零、开发环境

1、eclipse+tomcat（tomcat插件已安装到eclipse中）。

一、开发servlet步骤

1、在eclipse中新建工程

File —> New —> Other —> Web —> Dynamic Web Project，填写ProjectName，工程便创建好了。第二个图为创建好的目录结构。

2、创建HTML文件

创建路径为 MyFirstServlet —> WebContent —> login.htm

下面是login.html代码

<html>
<head>
<title>sayYourName</title>
</head>
<body>
<form name = "loginForm" method = "POST" action = "dispatcher">
<table>
<tr>
<td><div align = "right">User Name:</div></td>
<td><input type = "text" name = "username"></td>
</tr>
<tr>
<td><div align = "right">Password:</div></td>
<td><input type = "password" name = "password"></td>
</tr>
<tr>
<td><input type = "submit" name = "submit" value = "submit"></td>
<td><input type = "reset" name = "reset" value = "reset"></td>
</tr>
</table>
</form>
</body>
</html>

3、创建servlet类

我创建的包名是mypack，类名是MyFirstServlet

创建路径为 MyFirstServlet —> src —> mypack —> MyFirstServlet.java

下面是 MyFirstServlet.java代码

package mypack; import java.io.IOException; import javax.servlet.GenericServlet;
import javax.servlet.RequestDispatcher;
import javax.servlet.ServletContext;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse; public class MyFirstServlet extends GenericServlet{
private String target = "/sayYourName.jsp";
/*响应客户端请求*/
public void service(ServletRequest request, ServletResponse response)
throws ServletException, IOException{
//读取表单用户名
String username = request.getParameter("username");
//读取pwd
String password = request.getParameter("password");
//在request对象中添加USER属性
request.setAttribute("USER", username);
//添加PASSWORD属性
request.setAttribute("PASSWORD", password); /*把请求转发给hello.jsp*/
ServletContext context = getServletContext();
RequestDispatcher dispatcher = context.getRequestDispatcher(target);
dispatcher.forward(request, response); }
}