对eToOne的热切共享关系执行联接获取时出现JPA ArgumentException错误

Question

我正面临着一个连接获取和渴望关系的问题。

实体C与抽象实体E具有反向关系

在执行简单的namedQuery时，如

SELECT a FROM A a WHERE a.key = :key

其中参数'key'是字符串类型，那么我没有问题。从检索到的实体A访问子实体B，按照其假定执行子请求。

SELECT a FROM A a JOIN FETCH a.entitiesB WHERE a.key = :key

javax.ejb.EJBTransactionRolledbackException: The transaction has been marked rollback only because the bean encountered a non-application exception :javax.ejb.EJBTransactionRolledbackException : The transaction has been marked rollback only because the bean encountered a non-application exception :org.apache.openjpa.persistence.ArgumentException : The specified parameter of type "class org.apache.openjpa.util.LongId" is not a valid query parameter.
at org.apache.openejb.core.ivm.BaseEjbProxyHandler.convertException(BaseEjbProxyHandler.java:345)
at org.apache.openejb.core.ivm.BaseEjbProxyHandler.invoke(BaseEjbProxyHandler.java:283)
... 
Caused by: <openjpa-2.2.0-r422266:1244990 nonfatal user error> org.apache.openjpa.persistence.ArgumentException: The specified parameter of type "class org.apache.openjpa.util.LongId" is not a valid query parameter.
FailedObject: SELECT relation FROM Relation relation JOIN FETCH relation.accounts WHERE relation.key = :key [java.lang.String]
at org.apache.openjpa.jdbc.sql.DBDictionary.setUnknown(DBDictionary.java:1458)
at org.apache.openjpa.jdbc.sql.SQLBuffer.setParameters(SQLBuffer.java:544)
at org.apache.openjpa.jdbc.sql.SQLBuffer.prepareStatement(SQLBuffer.java:453)
at org.apache.openjpa.jdbc.sql.SQLBuffer.prepareStatement(SQLBuffer.java:429)
at org.apache.openjpa.jdbc.sql.SelectImpl.prepareStatement(SelectImpl.java:479)
... 

这是说明问题的代码示例。

实体A

@Entity
@Table(name = "TABLE_A")
@Access(AccessType.FIELD)
@NamedQueries({
   @NamedQuery(name = "findADetails", query = "SELECT a FROM A a WHERE a.customKey.key     = :customKey")})
public class A extends abstractEntity {

   @Embedded
   private CustomEmbeddedKey customKey;

   @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "entityA")
   private List<B> bEntities;

   ...
}

实体B

@Entity
@Table(name = "TABLE_B")
@Access(AccessType.FIELD)
public class B extends abstractEntity {

   @ManyToOne(fetch = FetchType.LAZY, targetEntity = A.class)
   @JoinColumn(name = "FK_A_ID")
   private A entityA;

   ...
}

@Entity
@Access(AccessType.FIELD)
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
@SequenceGenerator(name = "TOTO_ID_SEQ", sequenceName = "TOTO_ID_SEQ", initialValue = 1, allocationSize = 1)
public abstract class abstractEntity {

   @Id
   @Column(name = "ID")
   @GeneratedValue(generator = "TOTO_ID_SEQ", strategy = GenerationType.SEQUENCE)
   private Long id;

   @OneToOne(mappedBy = "...", fetch = FetchType.EAGER)
   private C anotherEntity;

   @OneToMany(mappedBy = "...", fetch = FetchType.LAZY)
   private List<D> anotherEntities;

   ...

}

@Embeddable
@Access(AccessType.FIELD)
public class CustomEmbeddedKey {

   @Column(name = "...", length = ...)
   private String key;

   ...
}

TypedQuery<A> query = createNamedQuery("findADetails", A.class);
  query.setParameter("customKey", queryParam);
  List<A> aEntitiesFound = query.getResultList();

Answer 1

我想我找到问题了。这是由于openJPA并不真正支持的映射，如果我们参考关于快速获取注意事项和限制的文档

最后，我有两个解决方案：

解决方案1（简单的一个）：

null

null

null

解决方案2：修改映射

• 从抽象实体中删除继承的用法（实际上没有业务用法），并用@mappedsuperclass替换（因此删除@entity和@inherity)
• 添加接口并定义热切属性的getters/setters
• 将此接口实现到抽象实体类中，以便子类必须实现getters/setters（以及相关的属性和映射）
• 将这些热切对象的目标实体的反向关系类型从抽象类类型更改为新接口类型
• 最后将EAGER属性的映射、getter和setter的实现从抽象实体类移到所有子实体类。

优点：

null

缺点：

• 更多修改
• 由于接口而不是抽象类类型到目标实体类中，对开发人员来说，可读性可能较低