Java 在JPA/Spring引导中重用实体

我有一个无法修改的现有实体，如下所示：

@Entity
public class ChildEntity implements Serializable{
private static final long serialVersionUID = 1L;
@Id
@Column(name="id")
private String id;
@Column(name="name")
private String name;
//getters, setters, et al.

@Entity
public class ParentEntity implements Serializable{
private static final long serialVersionUID = 1L;
@Column(name="extra")
private String extra;
@Id
@Column(name="id")
private String id;
@Column(name="name")
private String name;
//getters, setters, et al. from child

我有一个新的查询，返回与上面相同的数据，并有一个额外的列，如下所示：

@Entity
public class ChildEntity implements Serializable{
private static final long serialVersionUID = 1L;
@Id
@Column(name="id")
private String id;
@Column(name="name")
private String name;
//getters, setters, et al.

@Entity
public class ParentEntity implements Serializable{
private static final long serialVersionUID = 1L;
@Column(name="extra")
private String extra;
@Id
@Column(name="id")
private String id;
@Column(name="name")
private String name;
//getters, setters, et al. from child

如上所述，我想将上述类简化为如下内容：

@Entity
public class ParentEntity implements Serializable{
private static final long serialVersionUID = 1L;
@Column(name="extra")
private String extra;
@Embedded
private ChildEntity entity; //ChildEntity will have annotation @Embeddable
//getters, setters, of extra param

ChildEntity: select name,id from table where (/*some condition*/)
ParentEntity: select extra,name,id from table where (/*some condition*/)

但是，当我运行本机查询将数据分配给

ParentEntity

列表时，它无法识别

ChildEntity

我的问题是这样的：

@Entity
public class ParentEntity implements Serializable{
private static final long serialVersionUID = 1L;
@Column(name="extra")
private String extra;
@Embedded
private ChildEntity entity; //ChildEntity will have annotation @Embeddable
//getters, setters, of extra param

ChildEntity: select name,id from table where (/*some condition*/)
ParentEntity: select extra,name,id from table where (/*some condition*/)

我该怎么做？

如果您阅读

可嵌入类用于表示实体的状态，但 与实体类不同，它们没有自己的持久标识

---

回顾您的实体，您可能需要

@MappedSuperclass

或

@heritation

，具体取决于您的模型关联。这个线程清楚地说明了这一点。

尝试添加整个查询创建以及赋值

实体

s不能

可嵌入

，它是或。