Question

在处理基于JSF的工作时，我遇到了Hibernate继承问题。起初，我的目标是使用由两个具体类实现的接口，但由于映射接口不可行，我使用了由两个具体类扩展的抽象类。在代码方面：

@Entity
@Inheritance(strategy=InheritanceType.JOINED)
public abstract class AbstractEntity implements Serializable {

    protected Long id;

    @Id
    @GeneratedValue(strategy=GenerationType.TABLE)
    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }
}

@Entity
@PrimaryKeyJoinColumn(name="id_company")
public class Company extends AbstractEntity {

 private Individal signer;
    private Individuo billHolder;
    private String partIva;

    @OneToOne
    public Individual getSigner() {
        return signer;
    }

    public void setSigner(Individual signer) {
        this.signer = signer;
    }

    @OneToOne
    public Individual getBillHolder() {
        return billHolder;
    }

    public void setBillHolder(Individual billHolder) {
        this.billHolder = billHolder;
    }

    public String getPartIva() {
        return partIva;
    }

    public void setPartIva(String partIva) {
        this.partIva = partIva;
    }
}

@Entity
@PrimaryKeyJoinColumn(name="id_private")
public class Private extends AbstractEntity {

 private Individual billHolder;

    @OneToOne
    public Individual getBillHolder() {
        return billHolder;
    }

    public void setBillHolder(Individual billHolder) {
        this.billHolder = billHolder;
    }

    @OneToOne
    public Individual getSigner() {
        return billHolder;
    }

    public void setSigner(Individual signer) {
    }
}

现在我有了第四个类（名为“Client”），它与AbstractEntity具有一对一的单向关系：

@Entity
public class Client implements Serializable {

    private Long id;
    private AbstractEntity holder;

    ...

    @OneToOne
    public SoggettoAstratto getHolder() {
        return holder;
    }

    public void setHolder(SoggettoAstratto cliente) {
        this.intestatario = cliente;
    }

   ...

而个人是一个描述性的阶级，没有关系和财产，其中包括CodFis。现在，如果我使用此条件进行查询：

Criteria crit = super.getSession().createCriteria(getPersistentClass());
crit.createCriteria("holder.billHolder")
   .add(Restrictions.eq("codFis", codFis));
List risultati = crit.list();

*NOTE: the code is part of a DAO. The getPersistentClass returns a Client.class, and it works perfectly with other methods in this DAO.*

我得到了这个例外：

org.postgresql.util.PSQLException: ERROR: table name "individual1_" specified more than once
 at org.postgresql.core.v3.QueryExecutorImpl.receiveErrorResponse(QueryExecutorImpl.java:2062)
 at org.postgresql.core.v3.QueryExecutorImpl.processResults(QueryExecutorImpl.java:1795)
 at org.postgresql.core.v3.QueryExecutorImpl.execute(QueryExecutorImpl.java:257)
 at org.postgresql.jdbc2.AbstractJdbc2Statement.execute(AbstractJdbc2Statement.java:479)
 at org.postgresql.jdbc2.AbstractJdbc2Statement.executeWithFlags(AbstractJdbc2Statement.java:367)
 at org.postgresql.jdbc2.AbstractJdbc2Statement.executeQuery(AbstractJdbc2Statement.java:271)
 at org.hibernate.jdbc.AbstractBatcher.getResultSet(AbstractBatcher.java:186)
 at org.hibernate.loader.Loader.getResultSet(Loader.java:1787)

我无法弄清楚问题是什么。谁做了？

提前致谢。 JLPicard

P.S。属性名称是从我的语言翻译而来的，因此它们的平均值与英语不完全相同。

涉及连接子类继承的一对一关联

0 个答案: