1
首先,抱歉我的英文不好。Hibernate OneToOne将列表添加到列
我是一名Java程序员,但我没有足够的背景来使用Hibernate。由于在办公室我们正在尝试学习一些新的技术,我正尝试使用Hibernate来帮助我们处理数据库问题。
我有两个班,一个叫Processamento和另一个叫菲拉
表processamento具有在菲拉一个注册表,由fila.id
Processamento类中定义
@Entity
@Table(name="ra_fila_processamento", schema = "processamento")
public class Processamento implements java.io.Serializable {
private static final long serialVersionUID = 1L;
private int idProcessamento;
private int idFila;
private int idServidor;
private Fila fila;
@Column(name="rfp_id")
public int getIdProcessamento() {
return idProcessamento;
}
public void setIdProcessamento(int idProcessamento) {
this.idProcessamento = idProcessamento;
}
@Id
@Column(name="rfp_rf_id")
public int getIdFila() {
return idFila;
}
public void setIdFila(int idFila) {
this.idFila = idFila;
}
@Column(name="rfp_ser_id")
public int getIdServidor() {
return idServidor;
}
public void setIdServidor(int idServidor) {
this.idServidor = idServidor;
}
@OneToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
@JoinTable(name="ra_fila", schema="processamento", joinColumns = {@JoinColumn(name="rf_id", unique = true)})
public Fila getFila() {
return fila;
}
public void setFila(Fila fila) {
this.fila = fila;
}
}
Fila类
@Entity
@Table(name="ra_fila", schema="processamento")
public class Fila implements java.io.Serializable {
private static final long serialVersionUID = 1L;
public Fila() {}
private int idFila;
private String arquivoFonte;
private String status;
@Id
@Column(name = "rf_id", unique = true, nullable = false)
@GeneratedValue
public int getIdFila() {
return idFila;
}
public void setIdFila(int idFila) {
this.idFila = idFila;
}
@Column(name="rf_arquivo_fonte")
public String getArquivoFonte() {
return arquivoFonte;
}
public void setArquivoFonte(String arquivoFonte) {
this.arquivoFonte = arquivoFonte;
}
@Column(name="rf_status")
public String getStatus() {
return status;
}
public void setStatus(String status) {
this.status = status;
}
}
的问题是,当我运行这段代码:
Session session = HibernateUtil.getSessionFactory().openSession();
session.beginTransaction();
try {
Query query = session.createQuery("from Processamento");
@SuppressWarnings("unchecked")
List<Fila> fila = query.list();
session.getTransaction().commit();
if (fila.size() == 0) return false;
} catch (HibernateException e) {
e.printStackTrace();
} finally {
session.flush();
session.close();
}
所创建的HQL是这一个:
select processame0_.rfp_rf_id as rfp1_3_, processame0_.rfp_id as rfp2_3_, processame0_.rfp_ser_id as rfp3_3_, processame0_1_.fila_rf_id as fila4_2_ from processamento.ra_fila_processamento processame0_ left outer join processamento.ra_fila processame0_1_ on processame0_.rfp_rf_id=processame0_1_.rf_id
在这个HQL错误是这一部分:
processame0_1_.fila_rf_id as fila4_2_
This fila_不应该在那里,我该怎么做这个关系OneToOne正确?我只想获取Processamento列并获得它的Fila。在Fila,Processamento只有一个结果。
谢谢。
最好的问候。
[解决]
主要错误是,我不应该使用JoinTable,JoinTable就像一个工会条款,我想。删除JoinTable并添加子句JoinColumn解决了我的问题,并且select可以触发另一个选择并使关系存在。
很好,固定它,但我偷了与名称错误相关的问题。也许关系设置错误或查询?谢谢 – 2012-04-27 17:09:01