Question

我有两个实体，我想通过多个列加入。这些列由两个实体共享的@Embeddable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Foo s（其中AnEmbeddableObject是Bar的唯一键}）。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
}

最后是AnEmbeddedObject类：

@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，模式规范化程度很低，这是一个限制，AnEmbeddedObject的字段在每个表中重复。

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

我已经尝试过标记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，Foo使用它们来引用Bar）以及两个实体中生成的ID。在Bar中，4列的集合必须是唯一的，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不会让你轻松做你想做的事情。来自Hibernate documentation：

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）。 （强调添加）

因此，如果您不愿意将AnEmbeddableObject标识为Bar，那么Hibernate将不会懒散，自动为您检索Bar。当然，您仍然可以使用HQL编写在AnEmbeddableObject上加入的查询，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。

Hibernate / JPA注释中的多列连接

3 个答案: