java - 如何将两个ID插入链接两个用户的表中？

Question

我想建立一个关注机制，我有以下 User 类，它也是 friend 类:

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

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    @Column(name = "USER_ID")
    private Long id;

    private String firstName;

    private String lastName;

    private String username;

    private String password;

    @Column(unique = true)
    private String email;

    @Column(unique = true)
    private String phoneNumber;

    private int age;

    private Role role;

    @OneToMany(mappedBy = "shareId")
    private List<Share> shares;

    @OneToMany(mappedBy = "trackId")
    private List<Track> tracks;

    //number of days of cooldown
    private int cooldown;

    private long numberOfReports;

    @ManyToMany(cascade={CascadeType.ALL})
    @JoinTable(name="FRIENDS",
            joinColumns={@JoinColumn(name="USER_ID")},
            inverseJoinColumns={@JoinColumn(name="FRIEND_ID")})
    private Set<User> friends = new HashSet<User>();

    @ManyToMany(mappedBy="friends")
    private Set<User> befriended = new HashSet<User>();

这段代码自动创建一个名为friends的表，其中包含用户user_id和friend_id之间的链接，现在当我想在用户之间创建链接时，如何告诉Hibernate这样做？用户服务:

        User followerUser = userRepository.findByUsername(followerUsername);
        User followedUser = userRepository.findByUsername(followedUsername);
        return userRepository.SOMETHING(followerUser.getId(), followedUser.getId);
    }

userRepository 中的方法应该是什么样的？

Answer 1

您应该能够通过更新用户并使用 save 来做到这一点，例如:

followerUser.getFriends().add(followedUser);
// This is not necessary for database, but useful on object level
followedUser.getBefriended().add(followerUser);

userRepository.save(followerUser);