持久化对象的可插入、可更新属性行为

Question

我有一个实体如下

@Entity
@Table(name="DATMLAUENDCTT")
public class EnderecoContatoLaudoEntity implements Serializable {

    /** A Constante serialVersionUID. */
    private static final long serialVersionUID = 1L;

    /** O id. */
    @EmbeddedId
    private EnderecoContatoLaudoEntityPK id;

    /** O nome contato. */
    @Column(name="CTTNOM")
    private String nomeContato;

    /** O Codigo ddd telefone contato. */
    @Column(name="CTTTELDDDCOD")
    private BigDecimal codigoDDDTelefoneContato;

    /** O numero telefone contato. */
    @Column(name="CTTTELNUM")
    private BigDecimal numeroTelefoneContato;

    //bi-directional many-to-one association to TipoTelefoneEntity
    /** O tipo telefone entity. */
    @ManyToOne
    @JoinColumn(name="TELTIPSEQNUM",referencedColumnName="TELTIPSEQNUM")
    private TipoTelefoneEntity tipoTelefoneEntity;

    //bi-directional many-to-one association to EnderecoLaudoEntity
    /** O endereco laudo entity. */
    @ManyToOne
    @JoinColumns({
        @JoinColumn(name="LAUSEQNUM", referencedColumnName="LAUSEQNUM"),
        @JoinColumn(name="ENDTIPSEQNUM", referencedColumnName="ENDTIPSEQNUM")
    })
    private EnderecoLaudoEntity enderecoLaudoEntity;

它的内嵌id如下：

@Embeddable
public class EnderecoContatoLaudoEntityPK implements Serializable {
    //default serial version id, required for serializable classes.
    /** A Constante serialVersionUID. */
    private static final long serialVersionUID = 1L;

    /** Coloquei os nomes das colunas que não vieram quando foi gerada pelo RAD. */
    @Column(name = "LAUSEQNUM", insertable = false, updatable = false)
    private long numeroSequenciaLaudo;

    /** O numero sequencia tipo endereco. */
    @Column(name = "ENDTIPSEQNUM", insertable = false, updatable = false)
    private long numeroSequenciaTipoEndereco;

    /** O numero sequencia tipo telefone. */
    @Column(name = "TELTIPSEQNUM", insertable = false, updatable = false)
    private long numeroSequenciaTipoTelefone;

我的问题是，为什么当我尝试持久化实体时插入不起作用？

错误告诉我 TELTIPSEQNUM 为空。 但我正在设置对象tipoTelefoneEntity，它对应于列TELTIPSEQNUM。

现在，当我去掉insertable = false，updatable = false，像这样：

/** O numero sequencia tipo telefone。 */ @Column(name = "TELTIPSEQNUM") private long numeroSequenciaTipoTelefone;

然后持久化工作。

插入不应该使用对象TipoTelefoneEntity中的TELTIPSEQNUM，即在嵌入的id对象之外吗？

Answer 1

简单，如果你使用

insertable = false

JPA 客户端不会在新对象中插入该值，应该如何工作 来自文档：

public abstract boolean insertable
(Optional) Whether the column is included in SQL INSERT statements generated by the persistence provider