通过单列而不是复合主键加入

Question

我有单个JPA实体,我想在这个表上添加自联接.表看起来像例如

@Entity
@Table(name = "TABLE_A")
@IdClass(TableAPk.class)
public class TableA implements Serializable {

    private static final long serialVersionUID = 1L;


    @Id
    @Column(name = "COLUMN_1", nullable = false, length = 64)
    private String            column_1;

    @Id
    @Column(name = "COLUMN_2", nullable = false, precision = 10, scale = 2)
    private BigDecimal        column_2;


    @ManyToOne
    @JoinColumn(name = "COLUMN_1", insertable = false, updatable = false)
    //@ManyToOne(optional = true, fetch = FetchType.LAZY)
    //@JoinTable(name = "KEY_MAPPING", 
    //        joinColumns = { @JoinColumn(name = "J_COLUMN_1", referencedColumnName = "COLUMN_1", insertable = false, updatable = false) } )
    private TableA tableA;

    @OneToMany(mappedBy="tableA", fetch = FetchType.LAZY)
    private Set<TableA> tableASet;

而IdClass看起来像:

public class TableAPk implements Serializable {
    // default serial version id, required for serializable classes.
    /** The Constant serialVersionUID. */
    private static final long serialVersionUID = 1L;

    private String            column_1;

    private BigDecimal        column_2;

根据我的业务逻辑,我需要在单列上添加自联接

final Join<TableA, TableA> joinASelf = joinX.join("tableA", JoinType.INNER);

但是表具有复合主键,因此使用@Id注释了多个字段.我得到例外:

由以下原因引起:org.hibernate.AnnotationException:从com.data.TableA引用com.data.TableA的外键具有错误的列数.应该是2.

如何在此处仅在单列上添加自联接？我是JPA的新手,所以如果我错过了什么,请告诉我.

2015年2月21日更新:我添加了注释@AssociationOverride来覆盖关联:

@ManyToOne
@AssociationOverride(name="tableA", 
        joinColumns=@JoinColumn(name="COLUMN_1"))
private TableA tableA;

生成的列名称显示为"TABLE_A_COLUMN_2".我无法找出原因.任何线索？

Answer 1

1. 正确的方法是(不是OP想要的)

你可以使用@JoinColumns注释(注意复数而不是单数),从而加入:

@Entity
@Table(name = "TABLE_A")
@IdClass(TableAPk.class)
public class TableA implements Serializable {

    private static final long serialVersionUID = 1L;


    @Id
    @Column(name = "COLUMN_1", nullable = false, length = 64)
    private String            column1;

    @Id
    @Column(name = "COLUMN_2", nullable = false, precision = 10, scale = 2)
    private BigDecimal        column2;


    @ManyToOne
    @JoinColumns({
        @JoinColumn(name = "FK_COL1", referencedColumnName="COLUMN_1"),
        @JoinColumn(name = "FK_COL2", referencedColumnName="COLUMN_2")
    })
    private TableA tableA;

    @OneToMany(mappedBy="tableA", fetch = FetchType.LAZY)
    private Set<TableA> tableASet;

2. 解决问题的方法:

删除这些@ManyToOne/ @OneToMany关系并添加一个private String refCol1;.有了它,为JPQL编写相应的Criteria Query:SELECT t1 FROM TableA t1, TableA t2 WHERE t2.refCol1 = t1.column1

随着CriteriyQuqeries解决方案看起来像:

final CriteriaBuilder criteriaBuilder = entityManagerMds.getCriteriaBuilder();
// This Pojo is used to fetch only selected fields
final CriteriaQuery<DummyPojo> createQuery = criteriaBuilder.createQuery(DummyPojo.class);

final Root<TableX> tableX = createQuery.from(TableX.class);
final Join<TableX, TableA> joinTableA = tableX.join("tableAs", JoinType.INNER);


// Every time you want to add a self join, create new root
final Root<TableA> tableA = createQuery.from(TableA.class);
final Predicate predicateSelfJoin = criteriaBuilder.equal(joinTableA.<String>get("column_1"), tableA.<String>get("column_1"));