Spring MVC错误

Question

我对计算机编程的世界是完全陌生的，所以要真正深入地理解许多概念是一件很困难的事情。现在，我正在做一个项目，在这个项目中我们实现了Spring MVC。该项目的第一步是使一个网站的登录页。我试着模仿我们在课堂上做的模型，但我似乎不能通过我的web浏览器中的以下错误：

Unsupported auto value type java.lang.String for field injuryReports.Login.userName

下面是我的Login实体类：

package injuryReports;

import java.io.Serializable;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;

@Entity
public class Login implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id @GeneratedValue
    private String userName;
    private String password;
    private int userId;

    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;
    }

    public int getUserId() {
        return userId;
    }

    public void setUserId(int userId) {
        this.userId = userId;
    }

    public Login() {
    }

    public Login(String userName, String password) {
        super();
        this.userName = userName;
        this.password = password;
    }

    public Login(int userId, String userName2, String password2) {
        this.userId = userId;
        this.userName = userName2;
        this.password = password2;
    }
}

我的LoginDao类：

package injuryReports;

import java.util.List;

import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.persistence.TypedQuery;
import org.springframework.stereotype.Component;
import org.springframework.transaction.annotation.Transactional;

/**
 * 
 * @author nGent
 *
 */

@Component
public class LoginDao {
    @PersistenceContext private EntityManager em;

    @Transactional
    public void persist(Login user) {
        em.persist(user);
    }

    public List<Login> getAllUsers() {
        TypedQuery<Login> query = em.createQuery(
                "Select u FROM Login u ORDER BY u.id", Login.class);
        return query.getResultList();
    }

    public Login validateLogin(String userName, String password) {
        Login login = null;
        TypedQuery<Login> query = em.createQuery(
                "Select u From Login u where u.userName = :userName " +
                " and u.password = :password", Login.class).setParameter(
                "userName", userName).setParameter("password", password);
        try {
            login = query.getSingleResult();
        }
        catch (Exception e) {
            //TODO: Handle Exception
        }
        return login;
    }
}

还有我的LoginController类：

package injuryReports;

import java.util.HashMap;
import java.util.Map;

import javax.servlet.http.HttpServletRequest;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.servlet.ModelAndView;

@Controller
public class LoginController {

    @Autowired
    private LoginDao loginDao;

    @RequestMapping(value = "/user", method = {RequestMethod.POST})
    public ModelAndView userEntry(HttpServletRequest request) {
        String userName = request.getParameter("userName");
        String password = request.getParameter("password");

        if (userName != "" && password != "") {
            loginDao.persist(new Login(userName, password));
        }

        return new ModelAndView("logon.jsp", "loginDao", loginDao);
    }

    @RequestMapping(value = "/login")
    public ModelAndView login(HttpServletRequest request) {
        String userName = request.getParameter("userName");
        String password = request.getParameter("password");
        String page = "login.jsp";

        if (userName != "" && password != "") {
            try {
                Login login = loginDao.validateLogin(userName, password);
                if (login != null) {
                    request.getSession().setAttribute("UserId", login.getUserId());
                    page = "login.jsp";
                }
            }
            catch (Exception e) {
                //TODO: Handle Exception
            }
        }
        return new ModelAndView(page, getDaos());
    }

    @RequestMapping(value = "/logon", method = {RequestMethod.GET})
    public ModelAndView logon(HttpServletRequest request) {
        //int userId = (Integer) request.getSession().getAttribute("userId");
        //request.getSession().setAttribute("UserID", userId);
        return new ModelAndView("logon.jsp", getDaos());
    }

    public Map<String, Object> getDaos() {
        Map<String, Object> models = new HashMap<String, Object>();
        models.put("loginDao", loginDao);
        return models;
    }
}

抱歉，这篇文章有点长--我想提供尽可能多的信息。我真的很感谢任何人的帮助！

Answer 1

不能对String属性使用@GeneratedValue。它根据底层数据库引擎使用数据库序列或AUTOINCREMENT特性。

删除此注释：

@Id
private String userName;

或者使用integer/long作为id：

@Id @GeneratedValue
private int userId;