我有两个表。一个是主键，另一个是复合键。
表A：
@Entity
@Table(name = "TableA")
public class TableA {
@Id
@Column(name = "myId")
private Long id;
@Column(name = "myName")
private String name;
@Column(name = "myRegion")
private String regionName;
}
表 B 的组合键：
@Embeddable
public class CompositePK implements Serializable {
private static final long serialVersionUID = 1L;
@Column(name = "myId", insertable=false, updatable=false)
private Long myId;
@Column(name = "secondaryId")
private String secondaryId;
}
表B：
@Entity
@Table(name = "TableB")
public class TableB {
@EmbeddedId
private CompositePK compositePK;
@Column(name = "Data")
private String regulationText;
}
现在我想实现查询
select * from TableA tableA
inner join TableB tableB
on tableA.myId=tableB.myId
where tableB.myId = 1;
我已经尝试了TableB中的以下片段(一对一连接)。
@OneToOne(cascade = CascadeType.ALL)
@JoinTable(name = "TableA", inverseJoinColumns = @JoinColumn(name = "myId", referencedColumnName = "myId"))
private TableA tableA;
但上面写着无效列secondaryId。为此奋斗一天。无法使用组合键中的主键联接表。任何帮助都将不胜感激。
应修改表 B 以使用“派生标识”：
@Entity
@Table(name = "TableB")
public class TableB {
@EmbeddedId
private CompositePK compositePK;
@ManyToOne(cascade = CascadeType.ALL)
@JoinColumn(name = "myId", referencedColumnName = "myId")
@MapsId("myId") // maps 'myId' attribute of embedded id CompositePK
private TableA tableA;
@Column(name = "Data")
private String regulationText;
}
衍生身份在第2.4.1节的JPA 2.1规范中进行了讨论（并举例说明）。
如果是复合键，您需要使用复合键对象访问它，如下所示：
@OneToOne(cascade = CascadeType.ALL)
@JoinTable(name = "TableA", inverseJoinColumns = @JoinColumn(name =
"compositePK.myId", referencedColumnName = "myId"))
private TableA tableA;
