Hibernate：使用在另一个实体中引用的复合键面对“id 的断列映射”

Question

我在 Hibernate 中创建一对一映射时遇到问题。以下是我试图实现它的方式。

以下是我的 SysEntity 超级类

@MappedSuperclass
public class BaseSysEntity {

@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private Integer id;

@Column(name="sysupdate")
private Date sysupdate;

public Integer getId() {
    return id;
}

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

public Date getSysupdate() {
    return sysupdate;
}

public void setSysupdate(Date sysupdate) {
    this.sysupdate = sysupdate;
}

}

下面的类是与“Project”创建oneToOne关系的实体

@Entity
@Table(name="sysproject")
public class SysProject extends BaseSysEntity implements Serializable {

@OneToOne(optional=true, fetch= FetchType.LAZY)
@PrimaryKeyJoinColumns({
    @PrimaryKeyJoinColumn(name="sysClientId", referencedColumnName="sysClientId"),
    @PrimaryKeyJoinColumn(name="pProject", referencedColumnName="pProject")
})
private Project project;

public Project getProject() {
    return project;
}

public void setProject(Project project) {
    this.project = project;
}
}

以下是项目类：

@Entity
@Table(name = "tproject")
public class Project {
    @EmbeddedId
    private ProjectID id; // It contains sysClientId and pProject as primary key
    @OneToOne(optional=true, mappedBy="project")
    private SysProject SysProject;

}

ProjectID 类：

@Embeddable
public class ProjectID implements Serializable{

@Column(name="pProject")
private String project;

@Column(name="sysClientId")
private String sysClientId;

public String getProject() {
    return project;
}

public ProjectID(){
    this.sysClientId="0";
}

public ProjectID(Integer number){
    this();
    this.project = number.toString();
}

public void setProject(String project) {
    this.project = project;
}

public String getSysClientId() {
    return sysClientId;
}

public void setSysClientId(String sysClientId) {
    this.sysClientId = sysClientId;
}
}

我得到的异常：

Caused by: org.hibernate.MappingException: broken column mapping for: SysProject.id of: com.spin.integration.dto.Project
at org.hibernate.persister.entity.AbstractPropertyMapping.initPropertyPaths(AbstractPropertyMapping.java:178)
at org.hibernate.persister.entity.AbstractPropertyMapping.initIdentifierPropertyPaths(AbstractPropertyMapping.java:249)
at org.hibernate.persister.entity.AbstractPropertyMapping.initPropertyPaths(AbstractPropertyMapping.java:222)
at org.hibernate.persister.entity.AbstractEntityPersister.initOrdinaryPropertyPaths(AbstractEntityPersister.java:2434)
at org.hibernate.persister.entity.AbstractEntityPersister.initPropertyPaths(AbstractEntityPersister.java:2471)
at org.hibernate.persister.entity.AbstractEntityPersister.postConstruct(AbstractEntityPersister.java:3766)
at org.hibernate.persister.entity.SingleTableEntityPersister.<init>(SingleTableEntityPersister.java:451)
at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:62)
at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
at java.lang.reflect.Constructor.newInstance(Constructor.java:408)
at org.hibernate.persister.internal.PersisterFactoryImpl.create(PersisterFactoryImpl.java:163)
at org.hibernate.persister.internal.PersisterFactoryImpl.createEntityPersister(PersisterFactoryImpl.java:135)
at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:386)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1790)
at org.hibernate.ejb.EntityManagerFactoryImpl.<init>(EntityManagerFactoryImpl.java:96)
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:914)
... 18 more

注意：这个 oneToOne 映射不存在于数据库级别（不幸的是，我无法对数据库进行更改）。

Answer 1

您可以检查此解决方案的工作情况： One Id of composite key to be referenced in another entity 他也面临同样的问题

@OneToOne(optional=true, fetch= FetchType.LAZY)
@JoinColumns({
    @JoinColumn(name="sysClientId", referencedColumnName="sysClientId"),
    @JoinColumn(name="pProject", referencedColumnName="pProject")
})

Answer 2

我不得不使用@JoinColumns，而不是使用@PrimaryKeyJoinColumns。