我正在开发一个JPA项目。我需要在具有三个主键的类上使用@OneToMany映射。您可以在此之后找到错误和类。
javax.persistence.PersistenceException: No Persistence provider for EntityManager named JTA_pacePersistence: Provider named oracle.toplink.essentials.PersistenceProvider threw unexpected exception at create EntityManagerFactory:
javax.persistence.PersistenceException
javax.persistence.PersistenceException: Exception [TOPLINK-28018] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.EntityManagerSetupException
Exception Description: predeploy for PersistenceUnit [JTA_pacePersistence] failed.
Internal Exception: Exception [TOPLINK-7220] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.ValidationException
Exception Description: The @JoinColumns on the annotated element [private java.util.Set isd.pacepersistence.common.Action.permissions] from the entity class [class isd.pacepersistence.common.Action] is incomplete. When the source entity class uses a composite primary key, a @JoinColumn must be specified for each join column using the @JoinColumns. Both the name and the referenceColumnName elements must be specified in each such @JoinColumn.
at oracle.toplink.essentials.internal.ejb.cmp3.EntityManagerSetupImpl.predeploy(EntityManagerSetupImpl.java:643)
at oracle.toplink.essentials.ejb.cmp3.EntityManagerFactoryProvider.createEntityManagerFactory(EntityManagerFactoryProvider.java:196)
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:110)
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:83)
at isd.pacepersistence.common.DataMapper.(Unknown Source)
at isd.pacepersistence.server.MainServlet.getDebugCase(Unknown Source)
at isd.pacepersistence.server.MainServlet.doGet(Unknown Source)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:718)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:831)
at org.apache.catalina.core.ApplicationFilterChain.servletService(ApplicationFilterChain.java:411)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:290)
at org.apache.catalina.core.StandardContextValve.invokeInternal(StandardContextValve.java:271)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:202)
以下是我的课程的源代码:
行动:
@Entity
@Table(name="action")
public class Action {
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private int num;
@ManyToOne(cascade= { CascadeType.PERSIST, CascadeType.MERGE,
CascadeType.REFRESH })
@JoinColumn(name="domain_num")
private Domain domain;
private String name;
private String description;
@OneToMany
@JoinTable(name="permission", joinColumns= { @JoinColumn(name="action_num", referencedColumnName="action_num", nullable=false, updatable=false) }, inverseJoinColumns= { @JoinColumn(name="num") })
private Set<Permission> permissions;
public Action() {
}
许可:
@SuppressWarnings("serial")
@Entity
@Table(name="permission")
public class Permission implements Serializable {
@EmbeddedId
private PermissionPK primaryKey;
@ManyToOne
@JoinColumn(name="action_num", insertable=false, updatable=false)
private Action action;
@ManyToOne
@JoinColumn(name="entity_num", insertable=false, updatable=false)
private isd.pacepersistence.common.Entity entity;
@ManyToOne
@JoinColumn(name="class_num", insertable=false, updatable=false)
private Clazz clazz;
private String kondition;
public Permission() {
}
PermissionPK:
@SuppressWarnings("serial")
@Entity
@Table(name="permission")
public class Permission implements Serializable {
@EmbeddedId
private PermissionPK primaryKey;
@ManyToOne
@JoinColumn(name="action_num", insertable=false, updatable=false)
private Action action;
@ManyToOne
@JoinColumn(name="entity_num", insertable=false, updatable=false)
private isd.pacepersistence.common.Entity entity;
@ManyToOne
@JoinColumn(name="class_num", insertable=false, updatable=false)
private Clazz clazz;
private String kondition;
public Permission() {
}
答案 0 :(得分:11)
早上好,
经过漫长的一天搜索JPA和@OneToMany如何使用复合PK,我确实找到了解决方案。为了使它工作,我使用了@OneToMany的参数mappedBY。正如您在代码示例中所看到的,我使用类Permission的属性action映射了Permission。就是这样!你知道的时候很简单!
FF
行动类:
@Entity
@Table(name="action")
public class Action {
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private int num;
@ManyToOne(cascade= { CascadeType.PERSIST, CascadeType.MERGE,
CascadeType.REFRESH })
@JoinColumn(name="domain_num")
private Domain domain;
private String name;
private String description;
@OneToMany(mappedBy="action")
private Set<Permission> permissions;
权限类
@SuppressWarnings("serial")
@Entity
@Table(name="permission")
public class Permission implements Serializable {
@EmbeddedId
private PermissionPK primaryKey;
@ManyToOne
@JoinColumn(name="action_num", insertable=false, updatable=false)
private Action action;
答案 1 :(得分:1)
错误消息似乎很清楚:您需要将复合PK的三列声明为@JoinColum
,并且必须为每个列指定name
和referenceColumnName
。我没有测试映射,但试试这个:
@OneToMany
@JoinTable(name="permission", joinColumns= {
@JoinColumn(name="col1", referencedColumnName="col1", nullable=false, updatable=false),
@JoinColumn(name="col2", referencedColumnName="col2", ...),
@JoinColumn(name="col3", referencedColumnName="col3", ...)
}, inverseJoinColumns= { @JoinColumn(name="num") })
private Set<Permission> permissions;