java - org.hibernate.TypeMismatchException : Provided id of the wrong type for class

Question

在spring mvc中，我想用hibernate将数据保存到mysql数据库中，并且还想通过电子邮件获取数据。在我的数据库中，我有表 users 表，其中包含以下字段:-(id(bigint),name(varchar),username(varchar),password(varchar))我有 Controller :

@RequestMapping(value = "/save", method = RequestMethod.POST)
    public String signup(@Valid @ModelAttribute("user") User user,HttpServletRequest request,
            HttpServletResponse response,BindingResult result) {

         userService.addUser(user.getName(),user.getUsername(),user.getEmail(),user.getPassword());


            java.util.List<User> user1== userService.getUserinfo(user.getEmail());

                      userService.display(user1);
}

UserDao.java 是:

public interface UserDao {

    public void saveUser ( User user );

     public  List<User> getUserinfo(String email);
}

UserDaoImpl 是:

@Transactional
@Repository("userDao")
public class UserDaoImpl implements UserDao {

public void saveUser(User user) {
        sessionFactory.getCurrentSession().save(user);
    }

public  List<User>  getUserinfo(String email) {

        String hql = "select * from users where email = :email";
    List<User> users = new ArrayList<User>();

    users = (List<User>) sessionFactory.getCurrentSession().createSQLQuery(hql)
            .setParameter("email", email).uniqueResult();

    return users;
    }

}

UserService.java 是:

public interface UserService {


    User addUser(String name, String username, String email, String password);

       public List<User> getUserinfo(String email);

}

UserServiceImpl 是:

 public class UserServiceImpl implements UserService {

    @Transactional(propagation = Propagation.REQUIRED)
        public User addUser(String name, String username, String email, String password) {


            User user = new User(name,username, email,password);

            userDao.saveUser(user);

            return user;
                }


  public List<User> getUserinfo(String email) {         
            return userDao.getUserinfo(email);
        }
    }

User.java 是:

     @Entity
        @Table(name = "users")
        public class User {

    @Column(name = "id")    
        private long userId;

    @Column(name = "name")  
            private String Name;

    @Column(name = "username")  
            private String username;

    @Column(name = "email") 
            private String email;

    @Column(name = "password")  
            private String password;
//constructors
        //setters and getters
        }

数据存储在数据库中，但是当我获取时 显示的错误是:

 com.news.User cannot be cast to java.util.List

Answer 1

异常消息非常清晰。 User 类的 ID 为 Long 类型，并且您尝试通过将其电子邮件作为 String 类型的参数传递来通过 ID 获取用户。这不可能行得通。

如果您的目标是通过电子邮件查找用户，那么您需要一个查询，例如

select u from User u where u.email = :email