Hibernate：手動調用SequenceGenerator？

Question

我寫了自己的IdGenerator：

public class AkteIdGenerator implements IdentifierGenerator {
   public Serializable generate(SessionImplementor session, Object object)
         throws HibernateException {
      // if custom id is set -> use this id
      if (object instanceof SomeBean) {
         SomeBean someBean = (SomeBean) object;
         Long customId = someBean.getCustomId();
         if (customId != 0) {
            return customId;
         }
      }
      // otherwise --> call the SequenceGenerator manually
      SequenceStyleGenerator sequenceGenerator ...
   }
}

有誰知道如何從我的生成器類調用sequenceGenerator，我通常可以根據注釋定義：

@GeneratedValue(
        strategy = GenerationType.SEQUENCE,
        generator = "MY_SEQUENCE")
@SequenceGenerator(
        allocationSize = 1,
        name = "MY_SEQUENCE",
        sequenceName = "MY_SEQUENCE_NAME")

我會非常感謝任何解決方案!!!!

非常感謝，諾伯特

Answer 1

您可以通過Generator類調用SequenceGenerator。 通過編寫此代碼。 自定義生成器類應該是

 public class StudentNoGenerator implements IdentifierGenerator {

public Serializable generate(SessionImplementor session, Object object)throws HibernateException {

    SequenceGenerator generator=new SequenceGenerator();
    Properties properties=new Properties();
    properties.put("sequence","Stud_NoSequence");
    generator.configure(Hibernate.STRING, properties, session.getFactory().getDialect());
    return generator.generate(session, session);

}

}
在上面的代碼中，Stud_NoSequence是可以創建的序列名稱。 在數據庫中通過wring create sequence Stud_NoSequence; Hibernate.String是SequenceGenerator類將返回的類型。

而域類將是

import javax.persistence.Column;
    import javax.persistence.Entity;
    import javax.persistence.GeneratedValue;
    import javax.persistence.Id;
    @Entity
    @org.hibernate.annotations.GenericGenerator(
    name = "Custom-generator",
    strategy = "com.ssis.id.StudentNoGenerator"
    )
    public class Student {
@Id @GeneratedValue(generator = "Custom-generator")
String rno;
@Column
String name;
public String getRno() {
    return rno;
}
public void setRno(String rno) {
    this.rno = rno;
}
public String getName() {
    return name;
}
public void setName(String name) {
    this.name = name;
}
    }

Answer 2

  @Id
  @GenericGenerator(name = "seq_id", strategy = "de.generator.AkteIdGenerator")
  @GeneratedValue(generator = "seq_id")
  @Column(name = "ID")
  private Integer Id;

http://blog.anorakgirl.co.uk/2009/01/custom-hibernate-sequence-generator-for-id-field/

Answer 3

不確定這是否有幫助，但我在搜索我的答案時一直看到這個帖子，我在任何地方找不到，但我自己找到了解決方案。 所以我認為這可能是分享的最佳場所。

如果您使用hibernate作為JPA提供程序，則可以手動調用分配給給定實體類的ID生成器。 首先注入JpaContext：

@Autowired
org.springframework.data.jpa.repository.JpaContext jpaContext;

然后獲取內部org.hibernate.id.IdentifierGenerator：

org.hibernate.engine.spi.SessionImplementor session = jpaContext.getEntityManagerByManagedType(MyEntity.class).unwrap(org.hibernate.engine.spi.SessionImplementor.class);
org.hibernate.id.IdentifierGenerator generator = session.getEntityPersister(null, new MyEntity()).getIdentifierGenerator();

現在，您可以以編程方式從生成器獲取ID：

Serializable id = generator.generate(session, new MyEntity());

Answer 4

您的帖子有助於更新序列的名稱。

因為我每月使用一個序列，並且配置不會更新每個標識符生成。

這是我的代碼：

@Override
public Serializable generate(SessionImplementor sessionImplementator,
        Object object) throws HibernateException {
    Calendar now = Calendar.getInstance();
    // If month sequence is wrong, then reconfigure.
    if (now.get(Calendar.MONTH) != SEQUENCE_DATE.get(Calendar.MONTH)) {
        super.configure(new LongType(), new Properties(),
                sessionImplementator.getFactory().getDialect());
    }
    Long id = (Long) super.generate(sessionImplementator, object);
    String sId = String.format("%1$ty%1$tm%2$06d", SEQUENCE_DATE, id);
    return Long.parseLong(sId);// 1301000001
}