Question

我有ID的课程

@org.hibernate.annotations.AccessType("property")
public class ObjectID implements java.io.Serializable
{   
private long value;  

    long getValue()
    {
       return value;
    }

    void setValue(Long id)
    {
       value = id != null ? id : 0L;
    }
  // equals, hash, contructor  
}

已映射使用此ID类的类（患者）。我想在ObjectID类中生成long值。帮帮我我试过了

public class Patient implements Serializable 
{
   @javax.persistence.Id
   @javax.persistence.Column(name = "aa_id")
   @org.hibernate.annotations.Formula("case when aa_id is null then patient_seq.nextval else aa_id end")
   @javax.persistence.AttributeOverride(name = "value", column = @Column(name = "aa_id"))
   private ObjectID id;
}

和

public class Patient implements Serializable 
{
   @javax.persistence.Id
   @javax.persistence.SequenceGenerator(name = "PatientSequenceGenerator",
                       sequenceName = "patient_seq")
   @javax.persistence.GeneratedValue(strategy = GenerationType.SEQUENCE,
                    generator = "PatientSequenceGenerator")
   @javax.persistence.AttributeOverride(name = "value", column = @Column(name = "aa_id"))
   private ObjectID id;
}

但是没有帮助

解决此问题的方法之一是为ObjectID编写自定义userType并编写自定义ID生成器。

Answer 1

compositeIds通常由程序分配，我不知道是否有可能通过开箱即用的数据库序列设置它（或部分）。

首先你可以尝试这样设置它是否有效：

public class Patient implements Serializable 
{
   @Id
   @SequenceGenerator(name = "PatientSequenceGenerator", sequenceName = "lab_patient_seq")
   @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "PatientSequenceGenerator")
   private Long id;
}

在override属性中生成值

1 个答案: