Question

我刚冬眠。我有两个具有一对多关系的表。两个表是：

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

这是一个父类，它与Image表具有一对多的关系

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;

现在，我要使用其父类的ID（即pashminaId）从imagenames类中选择一个Image

例如：

从TBL_IMAGE中选择IMAGE_NAME，其中PASHMINA_ID ='some_digit';

我如何在图像类中传递pashminaId，因为没有pashminaId，它只有父类Pashmina的对象创建。

那么，我如何在休眠状态下实现这一目标？

如果您不了解我，请告诉我。

谢谢！希望能得到积极的回应。

Answer 1

当Hibernate处理对象而不是SQL表时，您可以简单地从pashmina对象访问image对象，并且在查询中将处理Java对象/ POJO，因此您可以对其进行访问通过Image.pashmina.pashminaId。

因此您可以编写以下查询：

String hql = "select I.imageName FROM Image I WHERE I.pashmina.pashminaId = 10";
Query query = session.createQuery(hql);
List results = query.list();

休眠从子表中选择查询

如果您不了解我，请告诉我。

1 个答案: