[英]How to reference a parent's id in a child's id with JPA/Hibernate?
给定一个表( MY_TABLE_A
),该表将在每次新插入时自动增加其ID(即数据库中的第一条记录的ID属性设置为1,第二条记录的ID属性设置为2,第三条记录的ID属性设置为3 )。 我在说的ID是表的主键。
我还有另一个表( MY_TABLE_B
),该表引用原始表的主键。 当我尝试将两者都持久保存到Oracle数据库时,我收到一个org.hibernate.id.IdentifierGenerationException: ids for this class must be manually assigned before calling save()
我要完成的工作:每当我将对象持久保存到MY_TABLE_A
,我都希望MY_TABLE_B
插入一个具有与MY_TABLE_A
相同的ID的对象,因为它是自动递增的(直到插入后,它才知道下一个值是什么)。 为了明确起见,表A中的一个ID在表B中应该只有一个匹配的ID
以下是我的代码的一些摘要:
的Firstclass:
@Entity
@Table(name = "MY_SCHEMA.MY_TABLE_A")
@Component
public class FirstClass implements Serializable {
@Id
@SequenceGenerator(name = "MY_SEQ", sequenceName = "MY_SCHEMA.MY_SEQ", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "MY_SEQ")
@Column(name = "MY_ID")
private Integer myId;
// more variables, getters/setters
}
二等:
@Entity
@Table(name = "MY_SCHEMA.MY_TABLE_B")
@SecondaryTable(name = "MY_SCHEMA.MY_TABLE_A", pkJoinColumns = @PrimaryKeyJoinColumn(name = "MY_ID", referencedColumnName = "MY_ID"))
@Component
public class SecondClass {
@Id
@Column(name = "MY_ID")
private Integer myId;
// more variables, getters/setters
}
我在其中为Oracle中的每个服务插入新条目的服务层代码段:
firstClassService.insert();
secondClassService.insert();
有关firstClassService的insert()
详细信息:
public void insert() {
FirstClass obj = new FirstClass();
getCurrentSession().persist(obj);
}
用于secondClassService的insert()
:
public void insert() {
SecondClass obj = new SecondClass();
getCurrentSession().persist(obj);
}
UPDATE
FirstClass现在的样子:
@Entity
@Table(name = "MY_SCHEMA.MY_TABLE_A")
@Component
public class FirstClass implements Serializable {
@Id
@SequenceGenerator(name = "MY_SEQ", sequenceName = "MY_SCHEMA.MY_SEQ", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "MY_SEQ")
@Column(name = "MY_ID")
@OneToOne(mappedBy = "myId")
private Integer myId;
}
二等:
@Entity
@Table(name = "MY_SCHEMA.MY_TABLE_B")
@SecondaryTable(name = "MY_SCHEMA.MY_TABLE_B", pkJoinColumns = @PrimaryKeyJoinColumn(name = "MY_ID", referencedColumnName = "MY_ID"))
@Component
public class SecondClass implements Serializable {
@Id
@JoinColumn(name = "MY_ID", referencedColumnName = "MY_ID")
@OneToOne
private Integer restRequestId;
}
映射应如下所示:
@Entity
@Table(name = "MY_SCHEMA.MY_TABLE_A")
@Component
public class FirstClass implements Serializable {
@Id
@SequenceGenerator(name = "MY_SEQ", sequenceName = "MY_SCHEMA.MY_SEQ", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "MY_SEQ")
@Column(name = "MY_ID")
private Long myId;
@OneToOne(mappedBy = "firstClass", cascade = CascadeType.ALL)
private SecondClass secondClass;
}
@Entity
@Table(name = "MY_SCHEMA.MY_TABLE_B")
@Component
public class SecondClass implements Serializable {
@Id
@JoinColumn(name = "MY_ID", referencedColumnName = "MY_ID")
@OneToOne
private FirstClass firstClass;
}
设置Cascade选项后,您只需调用保存firstClass的方法:关联的secondClass将自动持久化-假设您在内存模型中设置了关联关系的两侧,即
firstClass.setSecondClass(secondClass);
secondClass.setFirstClass(firstClass);
将@GeneratedValue(strategy=GenerationType.IDENTITY)
添加到第二类的ID。
@Id
@Column(name = "MY_ID")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Integer myId;
// more variables, getters/setters
从您的描述看来,您似乎具有一个ManytoOne关系,因为您的表B引用了表A,因此逻辑上说A有一些B的列表,因此为什么不利用ORM的实际含义,又为什么不保留引用呢?如:
@OneToMany(mappedBy="aa")
private List<B> bs;
并在另一个实体中使用注释:
@ManyToOne
@JoinColumn(name = "myId" , referencedColumnName = "id")
private A aa;
结合Jens的建议,请参见OracleDialect不支持身份密钥生成
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