JPA，Composite Key 由外键和表列成员组成

Posted 2023-04-14

【中文标题】JPA，Composite Key 由外键和表列成员组成

【英文标题】：JPA, Composite Key consisted of foreign key and table column members

【发布时间】：2018-08-08 13:47:31

【问题描述】：

向社区致以问候，

我整天都在努力寻找解决以下问题的方法。

场景如下，我有一张桌子

---TABLE_ONE--- INT ID VARCHAR NAME PRIMARY_KEY (ID)

我的另一个表由三列组成，它们共同组成一个复合键

---TABLE_TWO--- INT TABLE_ONE_ID (FK -> TABLE_ONE.ID) VARCHAR NAME VARCHAR EMAIL PRIMARY_KEY(TABLE_ONE_ID, NAME, EMAIL)

我想要实现的关系是TABLE_ONE 实体将 拥有来自TABLE_TWO 的对象列表（一对多关系）。

我尝试这样做，如下所示。

@Entity
@Table(name = "TABLE_ONE")
public class TableOne 

  @Column(name="id")
  private int id;
  @Column(name="name")
  private String name
  @OneToMany(fetch = FetchType.EAGER, mappedBy = "tableOne")
  private List<TableTwo> tableTwoList;
  //getters, setters, constructors        


@Entity
@Table(name = "TABLE_TWO")
public class TableTwo 

  @EmbeddedId
  private TableTwoCompositeId tableTwoCompositeId;
  @ManyToOne
  @JoinColumn(name = "TABLE_ONE_ID", referencedColumnName = "ID", insertable = false, updatable = false)
  private TableOne tableOne;
  //getters, setters, constructors        



@Embeddable
public class TableTwoCompositeId 
    @Column(name = "TABLE_ONE_ID")
    public Integer provider;
    @Column(name = "NAME")
    public String name;
    @Column(name = "EMAIL")
    public String email;
    //getters, setters, constructors

但是，当从数据库中检索 TableOne 对象时，我得到了 javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not extract ResultSet 和 Caused by: java.sql.SQLSyntaxErrorException: ORA-00942: table or view does not exist。

提前感谢您的帮助！

【参考方案1】：

我认为您需要进行一些小改动：

TableOne.id 需要一个 @Id 注释 TableTwoCompositeId.provider 的类型应该与TableOne.id 的类型匹配 TableTwo.tableOne 需要一个 @MapsId 注释来指示它映射 TableTwoCompositeId.provider

以下是代码的外观：

@Entity
@Table(name = "TABLE_ONE")
public class TableOne 

  @Id
  @Column(name="id")
  private int id;
  @Column(name="name")
  private String name
  @OneToMany(fetch = FetchType.EAGER, mappedBy = "tableOne")
  private List<TableTwo> tableTwoList;
  //getters, setters, constructors        


@Entity
@Table(name = "TABLE_TWO")
public class TableTwo 

  @EmbeddedId
  private TableTwoCompositeId tableTwoCompositeId;
  @MapsId("provider") // maps provider attribute of embedded id
  @ManyToOne
  @JoinColumn(name = "TABLE_ONE_ID", referencedColumnName = "ID", insertable = false, updatable = false)
  private TableOne tableOne;
  //getters, setters, constructors        



@Embeddable
public class TableTwoCompositeId 
    @Column(name = "TABLE_ONE_ID")
    public int provider;
    @Column(name = "NAME")
    public String name;
    @Column(name = "EMAIL")
    public String email;
    //getters, setters, constructors

【讨论】：

感谢您的回复。我做了您建议的更改，但 List tableTwoList 的值低于Unable to evaluate the expression Method threw 'org.hibernate.exception.SQLGrammarException' exception.

听起来有点不对劲。也许您可以在问题中包含整个堆栈跟踪。