Java 复合密钥部分的JPA连接
我尝试用JPA连接两个表 第一个表与实体ReportTripSingle关联 第二个表与实体TripData关联 第二个表的主键用复合键(实体TripDataPK)描述 实体TripDataPK 实体数据 实体报告单 正如你所看到的,我想加入ReportTripSingle和TripData 它不起作用:( 以下是堆栈跟踪: 原因:org.hibernate.AnnotationException: 的引用列名称(FTP\U ID) com.nexo.susan.be.model.ReportTripSingle.tripData引用 com.nexo.susan.be.model.TripData未映射到 org.hibernate.cfg.BinderHelper.createSyntheticPropertyReference(BinderHelper.java:336) 在 org.hibernate.cfg.ToOneFkSecondPass.doSecondPass(ToOneFkSecondPass.java:116) 在 org.hibernate.cfg.Configuration.processEndOfQueue(Configuration.java:1522) 在 org.hibernate.cfg.Configuration.processFkSecondPassInOrder(Configuration.java:1443) 在 secondPassCompile(Configuration.java:1346) 在 org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1737) 事实上,TripData并没有映射到单个属性,但我想加入FTP\u IDJava 复合密钥部分的JPA连接,java,jpa,join,composite,Java,Jpa,Join,Composite,我尝试用JPA连接两个表 第一个表与实体ReportTripSingle关联 第二个表与实体TripData关联 第二个表的主键用复合键(实体TripDataPK)描述 实体TripDataPK 实体数据 实体报告单 正如你所看到的,我想加入ReportTripSingle和TripData 它不起作用:( 以下是堆栈跟踪: 原因:org.hibernate.AnnotationException: 的引用列名称(FTP\U ID) com.nexo.susan.be.model.ReportT
问候。我无法加入复合密钥的一部分。 为了解决这个问题,我修改了与实体ReportTripSingle关联的表,添加了ftpBatchId和ftpDay作为外键 如果有人感兴趣,以下是我的实现: 实体报告单
public class ReportTripSingle implements java.io.Serializable {
@Id
@Column(name = "TRE_ID", precision = 10, scale = 0)
private long id;
@Column(name = "TRE_FTP_ID", length = 15, nullable = false, precision = 10, scale = 0, insertable = false, updatable = false)
private long tripDataId;
@Column(name = "TRE_BATCH_ID", nullable = false, insertable = false, updatable = false)
private long tripBatchId;
@Column(name = "TRE_FTP_DEP_DAY", nullable = false,insertable = false, updatable = false)
private long tripDay;
@ManyToOne(fetch = FetchType.EAGER)
@JoinColumns({
@JoinColumn(name = "TRE_FTP_ID", referencedColumnName = "FTP_ID"),
@JoinColumn(name = "TRE_BATCH_ID", referencedColumnName = "FTP_BATCH_ID"),
@JoinColumn(name = "TRE_FTP_DEP_DAY", referencedColumnName = "FTP_DAY")
})
private TripData tripData;
}
更好的选择是在@Joincolumn中使用修饰符insertable=false和updateable=false,如中所述:
“它不工作”是什么意思?你能提供一些错误日志、异常、不希望出现的行为吗?我已经复制了这个bug并添加了stacktrace
public class TripData implements java.io.Serializable {
@EmbeddedId
private TripDataPK tripDataPK;
// some properties ...
}
public class ReportTripSingle implements java.io.Serializable {
@Id
@Column(name = "TRE_ID", precision = 10, scale = 0)
private long id;
@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name = "TRE_FTP_ID", referencedColumnName = "FTP_ID")
private TripData tripData;
}
public class ReportTripSingle implements java.io.Serializable {
@Id
@Column(name = "TRE_ID", precision = 10, scale = 0)
private long id;
@Column(name = "TRE_FTP_ID", length = 15, nullable = false, precision = 10, scale = 0, insertable = false, updatable = false)
private long tripDataId;
@Column(name = "TRE_BATCH_ID", nullable = false, insertable = false, updatable = false)
private long tripBatchId;
@Column(name = "TRE_FTP_DEP_DAY", nullable = false,insertable = false, updatable = false)
private long tripDay;
@ManyToOne(fetch = FetchType.EAGER)
@JoinColumns({
@JoinColumn(name = "TRE_FTP_ID", referencedColumnName = "FTP_ID"),
@JoinColumn(name = "TRE_BATCH_ID", referencedColumnName = "FTP_BATCH_ID"),
@JoinColumn(name = "TRE_FTP_DEP_DAY", referencedColumnName = "FTP_DAY")
})
private TripData tripData;
}
public class ReportTripSingle implements java.io.Serializable {
@Id
@Column(name = "TRE_ID", precision = 10, scale = 0)
private long id;
@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name = "TRE_FTP_ID", referencedColumnName = "FTP_ID", **insertable = false, updatable = false**)
private TripData tripData;
}