Hibernate不插入外键值,但提供相关实体
我有3个相互关联的类:parametrosential、nomficherosential和tipocamposetity。当试图持久化一个ParameterSensity对象,并为其提供另外两个所需的对象时,就会出现问题 参数意识:Hibernate不插入外键值,但提供相关实体,hibernate,hibernate-mapping,Hibernate,Hibernate Mapping,我有3个相互关联的类:parametrosential、nomficherosential和tipocamposetity。当试图持久化一个ParameterSensity对象,并为其提供另外两个所需的对象时,就会出现问题 参数意识: @Entity @Table(name = "CON_DPARAMETROS") public class ParametrosEntity { @EmbeddedId @AttributeOverrides({ @AttributeOverri
@Entity
@Table(name = "CON_DPARAMETROS")
public class ParametrosEntity {
@EmbeddedId
@AttributeOverrides({
@AttributeOverride(name = "componenteId", column = @Column(name = "COM_IDENTIFICADOR",
nullable = false)),
@AttributeOverride(name = "clave", column = @Column(name = "PAR_CLAVE",
nullable = false)),
@AttributeOverride(name = "fichero", column = @Column(name = "NOM_FICHERO",
nullable = false))})
private ParametrosPK id;
@Column(name = "PAR_VALOR")
private String valor;
@Column(name = "PAR_LABEL")
private String label;
@Column(name = "PAR_DESCRIPCION")
private String descripcion;
@Column(name = "PAR_PRESENTACION")
private String presentacion;
@Column(name = "PAR_PROGRAMA")
private String programa;
@OneToOne
@JoinColumns({
@JoinColumn(name = "COM_IDENTIFICADOR", referencedColumnName = "COM_IDENTIFICADOR",
nullable = false, insertable = false, updatable = false),
@JoinColumn(name = "NOM_FICHERO", referencedColumnName = "NOM_FICHERO",
nullable = false, insertable = false, updatable = false)})
private NombreFicherosEntity fichero;
@OneToOne
@JoinColumn(name = "TIP_TIPO", nullable = false, insertable = false, updatable = false)
private TipoCamposEntity tipo;
@Entity
@Table(name = "CON_DNOMFICHEROS")
public class NombreFicherosEntity {
@EmbeddedId
@AttributeOverrides({
@AttributeOverride(name = "idComponente", column = @Column(name = "COM_IDENTIFICADOR",
nullable = false)),
@AttributeOverride(name = "fichero", column = @Column(name = "NOM_FICHERO",
nullable = false))})
private NombreFicherosPK id;
@OneToOne
@JoinColumn(name = "COM_IDENTIFICADOR", nullable = false, insertable = false, updatable = false)
private ComponentesEntity componente;
TipoCamposEntity:
@Entity
@Table(name = "CON_PTIPOCAMPOS")
public class TipoCamposEntity {
@Id
@Column(name = "TIP_TIPO")
private String tipo;
@Column(name = "TIP_DESCRIPCION")
private String descripcion;
法律意识:
@Entity
@Table(name = "CON_DPARAMETROS")
public class ParametrosEntity {
@EmbeddedId
@AttributeOverrides({
@AttributeOverride(name = "componenteId", column = @Column(name = "COM_IDENTIFICADOR",
nullable = false)),
@AttributeOverride(name = "clave", column = @Column(name = "PAR_CLAVE",
nullable = false)),
@AttributeOverride(name = "fichero", column = @Column(name = "NOM_FICHERO",
nullable = false))})
private ParametrosPK id;
@Column(name = "PAR_VALOR")
private String valor;
@Column(name = "PAR_LABEL")
private String label;
@Column(name = "PAR_DESCRIPCION")
private String descripcion;
@Column(name = "PAR_PRESENTACION")
private String presentacion;
@Column(name = "PAR_PROGRAMA")
private String programa;
@OneToOne
@JoinColumns({
@JoinColumn(name = "COM_IDENTIFICADOR", referencedColumnName = "COM_IDENTIFICADOR",
nullable = false, insertable = false, updatable = false),
@JoinColumn(name = "NOM_FICHERO", referencedColumnName = "NOM_FICHERO",
nullable = false, insertable = false, updatable = false)})
private NombreFicherosEntity fichero;
@OneToOne
@JoinColumn(name = "TIP_TIPO", nullable = false, insertable = false, updatable = false)
private TipoCamposEntity tipo;
@Entity
@Table(name = "CON_DNOMFICHEROS")
public class NombreFicherosEntity {
@EmbeddedId
@AttributeOverrides({
@AttributeOverride(name = "idComponente", column = @Column(name = "COM_IDENTIFICADOR",
nullable = false)),
@AttributeOverride(name = "fichero", column = @Column(name = "NOM_FICHERO",
nullable = false))})
private NombreFicherosPK id;
@OneToOne
@JoinColumn(name = "COM_IDENTIFICADOR", nullable = false, insertable = false, updatable = false)
private ComponentesEntity componente;
我在尝试持久化ParameterSensity对象时遇到问题。我将数据分配到对象参数中,包括所需的实体,但插入时会出错
以下是我用于persist的代码和获得的错误:
持久化代码:
ParameterSentity param=新的ParameterSentity()
以及错误的输出:
Hibernate: insert into CON_DPARAMETROS (PAR_DESCRIPCION, PAR_LABEL, PAR_PRESENTACION, PAR_PROGRAMA, PAR_VALOR, PAR_CLAVE, COM_IDENTIFICADOR, NOM_FICHERO) values (?, ?, ?, ?, ?, ?, ?, ?)
ORA-01400: cannot insert NULL into ("SGAINSDIST"."CON_DPARAMETROS"."TIP_TIPO")
我可以看到hibernate在insert查询中没有包含“TIP_TIPO”列,但我不理解为什么没有。我做错了什么?谢谢你的帮助。你能展示一下如何创建
TipoCamposEntity
的对象吗?你试过这样做吗?:“@OneToOne@JoinColumn(name=“TIP\u TIPO”,nullable=false,insertable=true,updateable=true)私有TipoCamposEntity TIPO;´