JPA 连接父/子表名称 @OneToMany

Lig*_*ard 3 jpa openjpa

我们正在尝试使用基本的 @OneToMany 关系:

@Entity
@Table(name = "PARENT_MESSAGE")
public class ParentMessage {

 @Id
 @Column(name = "PARENT_ID")
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 private Integer parentId;

 @OneToMany(mappedBy="parentMsg",fetch=FetchType.LAZY)
 private List childMessages;

 public List getChildMessages() {
  return this.childMessages;
 }
 ...
}

@Entity
@Table(name = "CHILD_MSG_USER_MAP")
public class ChildMessage {

 @Id
 @Column(name = "CHILD_ID")
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 private Integer childId;

 @ManyToOne(optional=false,targetEntity=ParentMessage.class,cascade={CascadeType.REFRESH}, fetch=FetchType.LAZY)
 @JoinColumn(name="PARENT_ID")
 private ParentMessage parentMsg;

 public ParentMessage getParentMsg() {
  return parentMsg;
 }
 ...
}
Run Code Online (Sandbox Code Playgroud)
   ChildMessage child = new ChildMessage();
   em.getTransaction().begin();
   ParentMessage parentMessage = (ParentMessage) em.find(ParentMessage.class, parentId);
   child.setParentMsg(parentMessage);
   List list = parentMessage.getChildMessages();
   if(list == null) list = new ArrayList<ChildMessage>();
   list.add(child);
   em.getTransaction().commit();
Run Code Online (Sandbox Code Playgroud)

我们收到以下错误。为什么 OpenJPA 将表名连接到APP.PARENT_MESSAGE_CHILD_MSG_USER_MAP?当然该表不存在..定义的表是APP.PARENT_MESSAGEAPP.CHILD_MSG_USER_MAP

引起原因:org.apache.openjpa.lib.jdbc.ReportingSQLException:表/视图“APP.PARENT_MESSAGE_CHILD_MSG_USER_MAP”不存在。{从 APP.PARENT_MESSAGE_CHILD_MSG_USER_MAP t0 内部联接 APP.CHILD_MSG_USER_MAP t1 ON t0.CHILDMESSAGES_CHILD_ID = t1.CHILD_ID WHERE t0.PARENTMESSAGE_ 选择 t1.CHILD_ID、t1.PARENT_ID、t1.CREATED_TIME、t1.USER_ID PARENT_ID = ?} [代码=30000,状态= 42X05]

Vin*_*nie 5

您可能想要将mappedBy 属性添加到关系的拥有方。这告诉 JPA 这只是一种关系,而不是两种不同的关系。也许从很多方面来说。