Question

我有一张桌子，桌子的结构是这样

package com.abc.domain;


@Entity
@Table(
    name = "ENTITY_DTL"
)

public class MyEntityDtl implements Serializable {

    private static final long serialVersionUID = 1L;
    private static final int NATLTY_CD_LENGTH = 3;


    @Id
    @Column(
        name = "MY_ENTITY_DTL_ID",
        unique = true,
        nullable = false
    )
    @GeneratedValue(
        strategy = GenerationType.SEQUENCE,
        generator = "myEntityDtlIdSeq"
    )
    @SequenceGenerator(
        name = "myEntityDtlIdSeq",
        sequenceName = "MY_ENTITY_DTL_ID_SEQ",
        allocationSize = 1
    )
    private Long MyEntityDtlId;



    @ManyToOne(
        fetch = FetchType.LAZY
    )
    @JoinColumns({@JoinColumn(
    name = "ENTITY_ID",
    referencedColumnName = "ENTITY_ID",
    nullable = false
), @JoinColumn(
    name = "ENTITY_TYPE_CD",
    referencedColumnName = "ENTITY_TYPE_CD",
    nullable = false
)})
    private MyEntity myEntity;


    /* 
    other columns and their getter setter 
    */


}

在这里-我要实现的是-像这样

使用此列-ENTITY_ID和ENTITY_TYPE_CD将MyEntityDtl表与Abc表相连。

如何在不使用查询中的MyEntity表的情况下实现它。

在这种情况下，这将使我失去更多的加入机会

感谢您的帮助，在此先感谢

Answer 1

由于访存类型是lazy，我直接使用join列来创建联接，因此不会影响查询性能，因为访存类型为lzay-

 Query query = getEntityManager().createQuery("SELECT abc FROM ABC abc, MyEntityDtl medtl join medtl.myEntity  me " +
                        " WHERE vsf.rqstId = me.entityId AND  " +
                        " me.entityTypeCd = '?' AND " +
                        " medtl.SOME_OTHER_COLUMN= '?' " ;
                query.setParameter("entityTypeCd", VALUE);
                query.setParameter("SOME_OTHER_COLUMN", VALUE);

Hibernate在where子句中使用联接列创建查询

1 个答案: