Hibernate Cascade来自另一个数据库的另一个Id

时间:2016-11-07 13:29:06

标签: java hibernate hibernate-mapping hibernate-annotations hibernate-cascade

例如,我有一个级联到B的A类:

public class A{

    private String id;
    private Set<B> bs = new HashSet<B>(0);

    @Id
    @GenericGenerator(name = "seq_id", strategy = generators.SequenceIdGenerator")
    @GeneratedValue(generator = "seq_id")
    @COLUMN(name="id" unique = true, nullable = false, length = 28)
    public void getId(){
        return this.id;
    }

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "A")
    @Cascade({CascadeType.SAVE_UPDATE, CascadeType.DELETE, CascadeType.PERSIST})
    public void getBs(){
        return bs;
    }
}

我有B,它有自己的id和一个包含来自A

的id的列
public class B

    String id; 
    String aId;
    A a;

    @Id
    @GenericGenerator(name = "seq_id", strategy = "generators.SequenceIdGenerator")
    @GeneratedValue(generator = "seq_id")
    @Column(name = "ID", unique = true, nullable = false, length = 28)
    public String getId() {
        return this.id;
    }

    @Column(name = "A_ID", nullable = false, length = 28)
    public String getAId() {
        return aId;
    }

    public void setAId(String aId) {
         this.aId = aId;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "A_ID", nullable = false, insertable = false, updatable = false)
    @MapsId("aId")
    public Admission getAdmission() {
        return this.admission;
    }

    public void setAdmission(Admission admission) {
    this.admission = admission;
    }

如何在保存A时自动分配aId?我想要设置aId,但我总是得到一个错误,表示A_ID不能设置为null。

谢谢!

1 个答案:

答案 0 :(得分:1)

在使用带有共享映射的派生标识符时,您不仅需要在关系中放置@MapsId anotation,还需要使用@Id注释来引用引用关系id的字段。

了解如何操作(setter ommited):

public class B {
    String id;
    String aId;
    A a;

    @Id
    @GenericGenerator(name = "seq_id", strategy = "generators.SequenceIdGenerator")
    @GeneratedValue(generator = "seq_id")
    @Column(name = "ID", unique = true, nullable = false, length = 28)
    public String getId() {
        return this.id;
    }

    @Id
    public String getAId() {
        return aId;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "A_ID", nullable = false)
    @MapsId("aId")
    public A getA() {
        return this.a;
    }
}

另一个错误是您将您的关系注释为updatable = falseinsertable = false,使持久性提供程序忽略将此字段写入数据库。