HQLHibernate不相关实体，但id相同

Question

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  public class Verification implements Serializable {
     private Long verificationId;
     private Long salesManCode;
     private Long clientCode;
     ...
  }

  public class Contract implements Serializable {
     private Long contractId;
     private Long salesManCode;
     private Long clientCode;
     ...
  }

这些类有各自的hbm.xml映射，在数据库模型中，这些表没有关联，在hibernate映射中也没有关联，但是在保存合同时，它必须在contractID字段（业务规则）中具有相同的verificationID，但在某些情况下，也给出了没有验证的合同。

verification
 verificationId | salesManCode | clientCode
  1050              1001            2056
  1051              1001            2248
  1054              1002            2856

contract
 contractId | salesManCode |clientCode
  1050         1001          2056         <- this contract have verification
  1051         1001          2248         <- this contract have verification
  1052         1025          2822         <- this contract not have verification
  1053         1254          1547         <- this contract not have verification
  1054         1002          2856         <- this contract have verification

我的问题是当我运行HQL查询时：

select con.salesManCode,  ver.salesManCode, con.clientCode, ver.clientCode, con.contracId, ver.verificationId 
from Verification ver, Contract con
where ver.verificationId = con.contractId

但Hibernate使用交叉连接语句翻译并合并所有记录。是否有方法在hbm.xml文件中的非相关类之间执行HQL查询？

规则：不应是映射的实体。

Answer 1

您可以按照以下方式映射两个表之间的联接关系：

@Entity
public class Verification implements Serializable {
    private Long verificationId;

    private Long salesManCode;
    private Long clientCode;
    ...
}

@Entity
public class Contract implements Serializable {
    private Long contractId;        

    @MapsId
    @OneToOne
    @JoinColumn(name = "contractId", referencedColumnName = "verificationId")
    private Verification verification;

    private Long salesManCode;
    private Long clientCode;
}

您的查询变为：

select con.salesManCode, ver.salesManCode, con.clientCode, ver.clientCode, con.contracId, ver.verificationId 
from Contract con
join con.verification ver

所以不是交叉连接，而是内部连接。