发布时间:2023-06-23 10:00
上一篇eclipse搭建maven web项目这里实现简单的servlet传参小Demo,项目结构:
1.pom.xml文件加入servlet包
4.0.0
com.citywy
mytest02
war
0.0.1-SNAPSHOT
junit
junit
3.8.1
test
javax.servlet
servlet-api
2.5
provided
mytest02
2.UserServlet.java
package com.citywy.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 UserServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public UserServlet() {
super();
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
doPost(request, response);
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
request.setCharacterEncoding("utf-8");
response.setContentType("text/html;charset=utf-8");
String name=request.getParameter("uname");
String pwd=request.getParameter("upwd");
System.out.println("输出"+name+"--"+pwd);
request.setAttribute("someinfo", "这是登陆成功提示信息!");
request.getRequestDispatcher("index.jsp").forward(request, response);
}
}
3.web.xml
Archetype Created Web Application
UserServlet
UserServlet
com.citywy.servlet.UserServlet
UserServlet
/UserServlet
4.index.jsp
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
User Login
用户登录
运行效果如图:
TIP:在Tomcat中运行时出现了无法加载到wtpwebapps或webapps下的情况,解决方案是servers下右击clean:将原先编译到tomcat服务器上的程序,先清除掉,然后再重新编译。