2

我正在尝试使用 Hibernate 连接到数据库。

此当前表格具有以下布局：

  (field1, field2, field3, optional1, optioanl2...)

其中field1、field2、field3都是外键，共同组成一个复合键。

我有以下课程：

@Entity
@Table(name = "db_table_mainRecords")
public class MainRecord implements Serializable{
    @EmbeddedId
    private MainRecordKey   lqk;
    @Transient
    private String          field1;
    @Transient
    private int         field2;
    @Transient
    private int         field3;
    @Column(name = "optional_1")
    private double          optional1;
    @Column(name = "optional_2")
    private double          optional2;

....
  // Getters and setters for all fields, including fields within MainClassKey
....
}

这伴随着它：

@Embeddable
@Table(name = "db_table_mainRecords")
public class MainRecordKey implements Serializable{
    @Column(name = "field1")
    private String field_1; 
    @Column(name = "field_2")
    private int field2;
    @Column(name = "field_3")
    private int field3;
}

我收到以下错误：

org.hibernate.QueryException - 无法解析属性：field3 of：org.hibernate.persister.entity.AbstractPropertyMapping.propertyException（AbstractPropertyMapping.java:67）的 path.MainRecord

这是我的博/道代码

  public List<MainRecord> getMainRecords(int e){
        Criterion crit1 = Restrictions.eq("field3", e);
        ArrayList<MainRecord> list = (ArrayList<MainRecord>) dao.getMainRecords(crit1);
}


public List<MainRecord> getMainRecords(Criterion criteria){
    if(criteria == null)
        return new ArrayList<MainReocrd>();
    return (List<MainRecord>) getHibernateTemplate().findByCriteria(DetachedCriteria.forClass(MainRecord.class).add(criteria));
}
4

1 回答 1

8

尝试int在映射文件Integer和. double_DoubleMainRecordMainRecordKey

更新 试试这个：

Restrictions.eq("lqk.field3", e);
于 2013-07-26T04:00:32.893 回答