用户/角色的系统,如何实现其他属性?

我想你应该把这些字段加进去 AppUser 在需要时初始化并插入数据。例如,创建时 Manager 您可以输入 admin_id 然后离开 client_id 空白的这就是 admin 也可以有多个 managers 在他下面。

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

    @Id @GeneratedValue
    private Long id;
    @Column(unique = true)
    private String username;
    private String password;
    @Column(unique = true)
    private String email;
    @Column(unique = true)
    private String phone;
    private String address;
    @ManyToMany(fetch = FetchType.EAGER)
    @JoinTable(name = "user_role", joinColumns = @JoinColumn(name = "user_id"), inverseJoinColumns = @JoinColumn(name = "role_id"))
    private Collection<AppRole> roles = new ArrayList<>();
    @ManyToOne
    @JoinColumn(name = "client_id")
    private AppUser client;
    @ManyToOne
    @JoinColumn(name = "manager_id")
    private AppUser manager;
    @ManyToOne
    @JoinColumn(name = "admin_id")
    private AppUser admin;
}

然后你可以分别给出 roles 致各 appUser 根据您的要求。 如果您尚未使用 Spring Security 然后,您应该将其用于角色和权限。