struts学习笔记(二)

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使用struts2建一个简单的登录页面。

首先配置struts.xml,如下：

<?xml version="1.0" encoding="UTF-8" ?>

<!DOCTYPE struts PUBLIC

"-//Apache Software Foundation//DTD Struts Configuration 2.0//EN"

"http://struts.apache.org/dtds/struts-2.0.dtd">

<result name="success">/loginsuccess.jsp</result>

<result name="input">/loginfail.jsp</result>

</action>

</package>

</struts>

写对应的action，在src目录下建一个包，名为com.login.action,在里面建立一个类

public class LoginAction extends ActionSupport{

User u;

UserService us = new UserService();

public User getU() {

return u;

}

public void setU(User u) {

this.u = u;

}

public UserService getUs() {

return us;

}

public void setUs(UserService us) {

this.us = us;

}

public String login(){

if(us.login(u))

return SUCCESS;

else

return INPUT;

}

Com.login.db下的类如下：

public class DB {

public static Connection getCon(){

Connection con=null;

try {

Class.forName("com.mysql.jdbc.Driver");

String url="jdbc:mysql://localhost/login?user=root&password=root";

con=DriverManager.getConnection(url);

} catch (ClassNotFoundException e) {

e.printStackTrace();

} catch (SQLException e) {

e.printStackTrace();

}

return con;

}

public static PreparedStatement prepare(Connection con,String sql){

PreparedStatement ps=null;

try {

ps=con.prepareStatement(sql);

} catch (SQLException e) {

e.printStackTrace();

}

return ps;

}

public static void close(Connection con){

try {

if(con != null) {

con.close();

con = null;

}

} catch (SQLException e) {

e.printStackTrace();

}

public static void close(Statement stmt){

try {

if(stmt != null) {

stmt.close();

stmt = null;

}

} catch (SQLException e) {

e.printStackTrace();

}

public static void close(ResultSet rs){

try {

if(rs != null) {

rs.close();

rs = null;

}

} catch (SQLException e) {

e.printStackTrace();

}

Com.login.model下的类如下：

public class User {

private String username;

private String password;

public String getUsername() {

return username;

}

public void setUsername(String username) {

this.username = username;

}

public String getPassword() {

return password;

}

public void setPassword(String password) {

this.password = password;

}

com.login.service下的类如下：

public class UserService {

public boolean login(User u){

Connection con = DB.getCon();

String sql = "select * from login where username='+u.getUsername()+' and password='+u.getPassword()+'";

PreparedStatement ps = DB.prepare(con, sql);

try {

ResultSet rs = ps.executeQuery();

if(rs.next())

return true;

} catch (SQLException e) {

e.printStackTrace();

}

return false;

}

相应的index.jsp文件如下：

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<html>

<head>

<title>My JSP 'index.jsp' starting page</title>

<!--

-->

</head>

<body>

username:<input name="u.username" type="text" size=15>

password:<input name="u.password" type="password" size=15>

</form>

</body>

</html>

loginfail.jsp如下：

<%@ page language="java" import="java.util.*" pageEncoding="ISO-8859-1"%>

String path = request.getContextPath();

String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<html>

<head>

<title>My JSP 'index.jsp' starting page</title>

<!--

-->

</head>

<body>

</body>

</html>

loginsuccess.jsp如下：

<%@ page language="java" import="java.util.*" pageEncoding="ISO-8859-1"%>

String path = request.getContextPath();

String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<html>

<head>

<title>My JSP 'index.jsp' starting page</title>

<!--

-->

</head>

<body>

</body>

</html>

如将以上文件拷贝到你的tomcat的webapps目录下面，部署好，如果不会配置，请看我的笔记一，然后启动你的tomcat。

如果登录不上，请读者自己修改db类中的数据库连接的语句，与你自己所在的机子的mysql密码，以及数据库想匹配上才能够正确登录。