Hibernate实体中的映射外键

Question

Hi编写Spring应用程序，使用Spring Security。这是我的用户和帐户角色数据库：

create table users (

  id int not null primary key,
  username varchar2(20) not null unique,
  password varchar2(20) not null,
  firstName varchar2(20),
  lastName varchar2(20),
  personalId varchar2(11) unique,
  city varchar2(40),
  address varchar2(40),
  email varchar2(30) unique,
  phone varchar2(9) unique,
  enabled number(1) not null
);

create table user_roles (
  id int primary key,
  name varchar2(20) not null,
  username varchar(20) constraint username_fk references users(username) not null
);

我的实体类：

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

    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue
    @Column(name = "id")
    private Integer id;
    @NotNull
    @Column(name = "username")
    private String username;
    @NotNull
    @Column(name = "password")
    private String password;
    @Column(name = "firstName")
    private String firstName;
    @Column(name = "lastName")
    private String lastName;
    @Column(name = "personalId")
    private String personalId;
    @Column(name = "city")
    private String city;
    @Column(name = "address")
    private String address;
    @Column(name = "email")
    private String email;
    @Column(name = "phone")
    private String phone;
    @NotNull
    @Column(name = "enabled")
    private int enabled;
    @OneToMany(mappedBy = "username")
    private Set<UserRole> userRoleSet = new HashSet<UserRole>(0);

@Entity
@Table(name = "user_roles")
public class UserRole implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue
    @Column(name = "id")
    private Integer id;
    @NotNull
    @Column(name = "name")
    private String name;
    @JoinColumn(name = "username")
    @ManyToOne(targetEntity = User.class)
    private String username;

当我尝试登录我的系统我有错误：

Hibernate：选择userrolese0_. username作为username3_1_0_，userrolese0_. id作为id1_0_0_，userrolese0_. id作为id1_0_1_，userrolese0_. name作为name2_0_1_，userrolese0_. username作为username3_0_1_user_rolesuserrolese0_其中userrolese0_. username=？警告：org.hibernate.engine.jdbc.spi.SqlExceptionHelper-SQL错误：1722，SQLState：42000错误：org.hibernate.engine.jdbc.spi.SqlExceptionHelper-ORA-01722：无效号码

我的类实现了UserDetailsService:

    package pl.piotr.ibank.service;

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.SimpleGrantedAuthority;
import org.springframework.security.core.userdetails.User;
import org.springframework.security.core.userdetails.UserDetails;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.core.userdetails.UsernameNotFoundException;
import org.springframework.stereotype.Service;
import org.springframework.transaction.annotation.Transactional;

import pl.piotr.ibank.daointerface.UserDao;
import pl.piotr.ibank.model.UserRole;

@Transactional(readOnly = true)
@Service("userDetailsService")
public class MyUserDetailsService implements UserDetailsService {

    @Autowired
    UserDao userDao;

    @Override
    public UserDetails loadUserByUsername(String username)
            throws UsernameNotFoundException {

        pl.piotr.ibank.model.User user = userDao.findByUsername(username);

        List<GrantedAuthority> authorities = buildUserAuthority(user
                .getUserRole());

        return buildUserForAuthentication(user, authorities);

    }

    private List<GrantedAuthority> buildUserAuthority(Set<UserRole> userRoles) {
        Set<GrantedAuthority> setAuths = new HashSet<GrantedAuthority>();

        for (UserRole userRole : userRoles) {
            setAuths.add(new SimpleGrantedAuthority(userRole.getName()));
        }

        List<GrantedAuthority> result = new ArrayList<GrantedAuthority>(
                setAuths);
        return result;
    }

    private UserDetails buildUserForAuthentication(
            pl.piotr.ibank.model.User user, List<GrantedAuthority> authorities) {
        return new User(user.getUsername(), user.getPassword(), true, true,
                true, true, authorities);
    }

    public UserDao getUserDao() {
        return userDao;
    }

    public void setUserDao(UserDao userDao) {
        this.userDao = userDao;
    }
}

我想是的，我为表映射的foregin键不好。例如，从用户查询＜code＞并从表返回用户，但当我尝试获取User_roles时，出现了上述错误。请检查我的地图是否正确。我正在使用Oracle数据库和Hibernate。

Answer 1

问题是，当你映射实体时，Hibernate希望外键是被引用实体的id，也就是说，你应该映射用户id而不是用户名。

此外，您的实体映射似乎是错误的：您使用了一个ManyToOne，目标实体是用户，但属性的类型是String。AFAIK Hibernate会尝试将用户分配到用户名，这将非常失败。

所以表格应该是这样的：

create table user_roles (
  id int primary key,
  name varchar2(20) not null,
  userid int constraint userid_fk references users(id) not null
);

UserRole中的映射应该是：

@JoinColumn(name = "userid")
@ManyToOne
private User user;

加上< code>User中的反向映射:

@OneToMany(mappedBy = "user")
private Set<UserRole> userRoleSet;

作为旁注，请记住< code>id在HQL中是一个特殊的关键字，即它总是引用一个实体的id。如果< code>id始终是用< code>@Id注释的唯一属性，那就没有问题，但是如果您改变这种情况，您可能会遇到查询选择错误数据甚至失败的问题。