spring3.1使用JDBC数据源datasource的异常有关问题

package com.model;public class User {    private int id;    private String name;    public int getId() {        return id;    }    public void setId(int id) {        this.id = id;    }    public String getName() {        return name;    }    public void setName(String name) {        this.name = name;    }}

package com.dao;import com.model.User;public interface UserDAO {    public void save(User u);}

package com.dao.impl;import java.sql.Connection;import java.sql.SQLException;import javax.annotation.Resource;import javax.sql.DataSource;import org.springframework.stereotype.Component;import com.dao.UserDAO;import com.model.User;@Component("userDAO")public class UserDAOImpl implements UserDAO {        private DataSource dataSource ;    public DataSource getDataSource() {        return dataSource;    }    @Resource(name="dataSource")    public void setDataSource(DataSource dataSource) {        this.dataSource = dataSource;    }    @Override    public void save(User u) {        try {            Connection conn = dataSource.getConnection();            conn.createStatement().executeUpdate("insert into user values(null,'lisi')");            conn.close();        } catch (SQLException e) {            // TODO Auto-generated catch block            e.printStackTrace();        }                System.out.println("A user saved!");            }}

package com.service;import javax.annotation.Resource;import org.springframework.stereotype.Component;import com.dao.UserDAO;import com.dao.impl.UserDAOImpl;import com.model.User;@Component("userService")public class UserService {    private UserDAO userDAO = new UserDAOImpl();    public UserDAO getUserDAO() {        return userDAO;    }    @Resource(name="userDAO")    public void setUserDAO(UserDAO userDAO) {        this.userDAO = userDAO;    }        public void add(User u) {        userDAO.save(u);    }}

package com.test;import org.junit.Test;import org.springframework.context.ApplicationContext;import org.springframework.context.support.ClassPathXmlApplicationContext;import com.model.User;import com.service.UserService;public class UserServiceTest {    @Test    public void testAdd() {        User u = new User();        ApplicationContext context = new ClassPathXmlApplicationContext(new String[]{"beans.xml"});        UserService userService = (UserService)context.getBean("userService");        userService.add(u);    }}