Hibernate/JPA注释中的多列连接

Question

我有两个实体,我想通过多个列加入.这些列@Embeddable由两个实体共享的对象共享.在下面的示例中,Foo只能有一个Bar但Bar可以有多个Foos(其中AnEmbeddableObject是唯一的键Bar).这是一个例子:

@Entity
@Table(name = "foo")
public class Foo {
    @Id
    @Column(name = "id")
    @GeneratedValue(generator = "seqGen")
    @SequenceGenerator(name = "seqGen", sequenceName = "FOO_ID_SEQ", allocationSize = 1)
    private Long id;
    @Embedded
    private AnEmbeddableObject anEmbeddableObject;
    @ManyToOne(targetEntity = Bar.class, fetch = FetchType.LAZY)
    @JoinColumns( {
        @JoinColumn(name = "column_1", referencedColumnName = "column_1"),
        @JoinColumn(name = "column_2", referencedColumnName = "column_2"),
        @JoinColumn(name = "column_3", referencedColumnName = "column_3"),
        @JoinColumn(name = "column_4", referencedColumnName = "column_4")
    })
    private Bar bar;

    // ... rest of class
}

和Bar类:

@Entity
@Table(name = "bar")
public class Bar {
    @Id
    @Column(name = "id")
    @GeneratedValue(generator = "seqGen")
    @SequenceGenerator(name = "seqGen", sequenceName = "BAR_ID_SEQ", allocationSize = 1)
    private Long id;
    @Embedded
    private AnEmbeddableObject anEmbeddableObject;

    // ... rest of class
}

最后AnEmbeddedObject课程:

@Embeddable
public class AnEmbeddedObject {
    @Column(name = "column_1")
    private Long column1;
    @Column(name = "column_2")
    private Long column2;
    @Column(name = "column_3")
    private Long column3;
    @Column(name = "column_4")
    private Long column4;

    // ... rest of class
}

显然,模式很难规范化,这是一个限制,AnEmbeddedObject在每个表中重复的字段.

我遇到的问题是当我尝试启动Hibernate时收到此错误:

org.hibernate.AnnotationException: referencedColumnNames(column_1, column_2, column_3, column_4) of Foo.bar referencing Bar not mapped to a single property

我试过标记JoinColumns不可插入和可更新,但没有运气.有没有办法用Hibernate/JPA注释来表达这个？

Answer 1

这对我有用.在我的情况下,2个表foo和boo必须基于3个不同的列加入.请注意在我的情况下,在boo中3个常用列不是主键

即,基于3个不同列的一对一映射

@Entity
@Table(name = "foo")
public class foo implements Serializable
{
    @Column(name="foocol1")
    private String foocol1;
    //add getter setter
    @Column(name="foocol2")
    private String foocol2;
    //add getter setter
    @Column(name="foocol3")
    private String foocol3;
    //add getter setter
    private Boo boo;
    private int id;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "brsitem_id", updatable = false)
    public int getId()
    {
        return this.id;
    }
    public void setId(int id)
    {
        this.id = id;
    }
    @OneToOne
    @JoinColumns(
    {
        @JoinColumn(updatable=false,insertable=false, name="foocol1", referencedColumnName="boocol1"),
        @JoinColumn(updatable=false,insertable=false, name="foocol2", referencedColumnName="boocol2"),
        @JoinColumn(updatable=false,insertable=false, name="foocol3", referencedColumnName="boocol3")
    }
    )
    public Boo getBoo()
    {
        return boo;
    }
    public void setBoo(Boo boo)
    {
        this.boo = boo;
    }
}





@Entity
@Table(name = "boo")
public class Boo implements Serializable
{
    private int id;
    @Column(name="boocol1")
    private String boocol1;
    //add getter setter
    @Column(name="boocol2")
    private String boocol2;
    //add getter setter
    @Column(name="boocol3")
    private String boocol3;
    //add getter setter
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "item_id", updatable = false)
    public int getId()
    {
        return id;
    }
    public void setId(int id)
    {
        this.id = id;
    }
}

Answer 2

如果这不起作用我就没有​​想法了.这样,您可以获得两个表中的4列(Bar拥有它们并Foo使用它们来引用Bar)以及两个实体中生成的ID.4列的集合必须是唯一的,Bar因此多对一关系不会成为多对多关系.

@Embeddable
public class AnEmbeddedObject
{
    @Column(name = "column_1")
    private Long column1;
    @Column(name = "column_2")
    private Long column2;
    @Column(name = "column_3")
    private Long column3;
    @Column(name = "column_4")
    private Long column4;
}

@Entity
public class Foo
{
    @Id
    @Column(name = "id")
    @GeneratedValue(generator = "seqGen")
    @SequenceGenerator(name = "seqGen", sequenceName = "FOO_ID_SEQ", allocationSize = 1)
    private Long id;
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumns({
        @JoinColumn(name = "column_1", referencedColumnName = "column_1"),
        @JoinColumn(name = "column_2", referencedColumnName = "column_2"),
        @JoinColumn(name = "column_3", referencedColumnName = "column_3"),
        @JoinColumn(name = "column_4", referencedColumnName = "column_4")
    })
    private Bar bar;
}

@Entity
@Table(uniqueConstraints = @UniqueConstraint(columnNames = {
    "column_1",
    "column_2",
    "column_3",
    "column_4"
}))
public class Bar
{
    @Id
    @Column(name = "id")
    @GeneratedValue(generator = "seqGen")
    @SequenceGenerator(name = "seqGen", sequenceName = "BAR_ID_SEQ", allocationSize = 1)
    private Long id;
    @Embedded
    private AnEmbeddedObject anEmbeddedObject;
}

Answer 3

Hibernate不会让你轻松地做你想做的事情.从Hibernate文档:

请注意,将referencedColumnName用于非主键列时,关联的类必须是Serializable.另请注意,非主键列的referencedColumnName必须映射到具有单个列的属性(其他情况可能不起作用). (重点补充)

因此,如果您不愿意AnEmbeddableObject为Bar 制作标识符,那么Hibernate不会懒散,自动为您检索Bar.当然,您仍然可以使用HQL编写加入的查询AnEmbeddableObject,但如果您坚持使用Bar的多列非主键,则会丢失自动提取和生命周期维护.