Question

我是冬眠/ jpa的新手，我面临如何正确建模嵌套实体的困难，我的要求是，用户->具有多个角色->每个角色具有许多权限

角色的每个权限都必须作为单独的行保留在表中。并且，需要为用户及其角色提供单独的表；同样，用户的每个角色都需要作为单独的行保留在表中。我写了以下内容；但这不允许我创建具有已定义角色的新用户...

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

    @Id
    @GeneratedValue(generator = "UUID")
    @GenericGenerator(name = "UUID",strategy = "org.hibernate.id.UUIDGenerator")
    public UUID id;

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

    @Column(name = "password")
    public String password;

    @OneToMany(cascade = CascadeType.ALL, mappedBy="user")
    @LazyCollection(LazyCollectionOption.FALSE)
    public List<UserRole> roles;
}

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

    @Id
    @GeneratedValue(generator = "UUID")
    @GenericGenerator(name = "UUID",strategy = "org.hibernate.id.UUIDGenerator" )
    public UUID id;

    @ManyToOne
    @JoinColumn(name="username", nullable=false, referencedColumnName = "username")
    public User user;

    @Column(name = "role_name")
    public String roleName;

    @OneToMany(fetch = FetchType.EAGER, cascade = {CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REFRESH})
    @JoinColumn(name = "role_name", referencedColumnName = "role_name")
    public List<RolePermission> permissions;

}

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

    @Id
    @Column(name = "role_name")
    public String roleName;

    @Id
    @Column(name = "permission")
    public String permission;
}

出现以下错误，

2019-03-06 21：25：22.643错误48896 --- [nio-8090-exec-3] o.h.engine.jdbc.spi.SqlExceptionHelper：错误：重复的键值违反唯一约束“ uk_40fvvy071dnqy9tywk6ei7f5r”详细信息：密钥（role_name）=（所有者）已经存在。

Answer 1

在正确建模UserRole之后，它可以正常工作，下面是我为UserRole创建的新实体类，

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

    @Id
    @GeneratedValue(generator = "UUID")
    @GenericGenerator(
            name = "UUID",
            strategy = "org.hibernate.id.UUIDGenerator"
    )
    public UUID id;

    @ManyToOne
    @JoinColumn(name="username", nullable=false, referencedColumnName = "username")
    public User user;

    @Column(name = "role_name")
    public String roleName;

    @ManyToMany
    @JoinTable(name = "roles_permissions",joinColumns = @JoinColumn(name = "role_name"))
    public List<RolePermission> permissions;

}

嵌套实体建模

1 个答案: