Question

我试图让Hibernate @OneToOne注释使用2个类，Hito和Portada。 Portada表具有Hito的外键，一个名为hito的int属性。我的实体看起来像这样：

日塔：

@Entity
@Table(name = "hito")
public class Hito implements Serializable {
    //...other attributes
    private Portada portada;

    //...getters and setters from other attributes

    @OneToOne(cascade = CascadeType.ALL, fetch = FetchType.EAGER, mappedBy = "hito")
    public Portada getPortada(){ return portada;}

    public void setPortada(Portada portada){ this.portada = portada;}
}

拉波尔塔达：

@Entity
@Table(name = "portada")
public class Portada {
    //...other attributes
    private Hito hito;

    //...getters and setters from other attributes
    @OneToOne(fetch = FetchType.EAGER, optional = false)
    @JoinColumn(name = "hito")
    public Hito getHito() {return hito;}

    public void setHito(Hito hito) {this.hito = hito;}

}

当我调用hito.getPortada（）时，我期待一个Portada对象，但它返回null。

有什么建议吗？先谢谢

Answer 1

我尝试用代码重现您的问题：

@MappedSuperclass
public abstract class BaseEntity {
    @Id @GeneratedValue
    private Long id;

    @Version
    private long version;

    public Long getId() {
        return id;
    }

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

    public long getVersion() {
        return version;
    }

    public void setVersion(long version) {
        this.version = version;
    }
}

@Entity
@Table(name = "portada")
public class Portada extends BaseEntity {
    //...other attributes
    @OneToOne(fetch = FetchType.EAGER, optional = false)
    @JoinColumn(name = "hito")
    private Hito hito;

    //...getters and setters from other attributes
    public Hito getHito() {return hito;}

    public void setHito(Hito hito) {this.hito = hito;}

}

@Entity
@Table(name = "hito")
public class Hito extends BaseEntity implements Serializable {
    @OneToOne(cascade = CascadeType.ALL, fetch = FetchType.EAGER, mappedBy = "hito")
    private Portada portada;


    public Portada getPortada(){ return portada;}

    public void setPortada(Portada portada){ this.portada = portada;}
}

// app:

            Portada p = new Portada();
            Hito h = new Hito();

            p.setHito(h);
            h.setPortada(p);

            entityManager.persist(h);
            entityManager.flush();
            entityManager.clear();

            Hito h2 = entityManager.find(Hito.class, h.getId());
            System.out.println(h2.getPortada().toString());

            tx.commit();

最后find生成的sql：

select
    hito0_.id as id1_4_0_,
    hito0_.version as version2_4_0_,
    portada1_.id as id1_7_1_,
    portada1_.version as version2_7_1_,
    portada1_.hito as hito3_7_1_ 
from
    hito hito0_ 
left outer join
    portada portada1_ 
        on hito0_.id=portada1_.hito 
where
    hito0_.id=?

一切都对我有用......

编辑：唯一的区别是我喜欢将字段属性放在字段而不是属性上，但在这个问题上并不重要。请检查是否将两个类添加到persistance.xml或hibernate config。

Hibernate @OneToOne无法正常工作

1 个答案: