我很冬眠。我有兩個表具有一對多的關系。兩張表是：public class Pashmina implements Serializable {    @Id    @GeneratedValue(strategy = GenerationType.AUTO, generator = "sq_pashmina_id")    @SequenceGenerator(name = "sq_pashmina_id", sequenceName = "sq_pashmina_id")    @Column(name = "PASHMINA_ID")    private int pashminaId;    @Column(name = "PASHMINA_NAME")    private String pashminaName;    @Column(name = "PRICE")    private double price;    @Column(name = "ADDED_AT", insertable = false)    @Temporal(TemporalType.TIMESTAMP)    private Date addedAt;    @Column(name = "CATEGORY")    private String category;    @Column(name = "ENABLED", insertable = false)    private Character enabled;    @OneToMany(mappedBy = "colourId", fetch = FetchType.EAGER)    private Set<PashminaColour> pashminaColor  = new HashSet<PashminaColour>();    @OneToMany(mappedBy = "imageId", fetch = FetchType.EAGER)    private Set<Image> images  = new HashSet<Image>();    @OneToMany(mappedBy = "descriptionId", fetch = FetchType.EAGER)    private Set<Description> descriptions  = new HashSet<Description>();    //getter and setter method這是一個父類，它與Image表具有一對多關系public class Image implements Serializable {    @Id    @Column(name = "IMAGE_ID")    private int imageId;    @Column(name = "IMAGE_NAME")    private String imageName;    @JoinColumn(name = "PASHMINA_ID", referencedColumnName = "PASHMINA_ID")    @ManyToOne    private Pashmina pashmina;現在，我要使用其父類的ID（即pashminaId）imagenames從Image類中選擇一個例如：從TBL_IMAGE中選擇IMAGE_NAME，其中PASHMINA_ID ='some_digit';我如何在圖像類中傳遞pashminaId，因為沒有pashminaId它只有Parent類的Object創建Pashmina。那么，我該如何在休眠狀態下實現呢？