Question

我尝试持久化一个与另一个子实体连接的父实体，但问题是在持久化时没有为这个子节点生成id，所以我有这个错误：title: This is the title author: - Author Juan$^1$ - Author Tu$^2$ institute: - $^1$Juans Casa - $^2$Tus Place date: "Thursday, April 09, 2015" output: beamer_presentation

有子实体：

[org.hibernate.engine.jdbc.spi.SqlExceptionHelper] ORA-01400: cannot insert NULL into ("L2S$OWNER"."SABRI"."TRANSITION_MATRIX_ID")

PK课程：

@Data
@Entity
@IdClass(MyLibrarySabriEntityPK.class)
@Table(name = "SABRI", schema = "L2S$OWNER", catalog = "")

public class MyLibrarySabriEntity extends ActionForm {
@Access(AccessType.FIELD)
@Id
@ManyToOne
@JoinColumn(name = "TRANSITION_MATRIX_ID", referencedColumnName = "ID_TRANSITION_MATRIX")
private MyLibraryTestEntity sabriEntity;

@Id
private String RATING_ID_ROW;

@Id
private String RATING_ID_COL;


@Basic
@Column(name = "TRANSITION_PROBABILITY", nullable = true, insertable = true, updatable = true, precision = 20)
private Double TRANSITION_PROBABILITY;}

}

有父实体：

@Data
public class MyLibrarySabriEntityPK implements Serializable {
private String TRANSITION_MATRIX_ID;
private String RATING_ID_ROW;
private String RATING_ID_COL;


public MyLibrarySabriEntityPK(String TRANSITION_MATRIX_ID,String RATING_ID_COL,String RATING_ID_ROW ){
    this.TRANSITION_MATRIX_ID=TRANSITION_MATRIX_ID;
    this.RATING_ID_COL = RATING_ID_COL;
    this.RATING_ID_ROW= RATING_ID_ROW;
}

在这里，我尝试保留父表，该表也应该持久保存子表，但不生成Id！

@Data
@Entity
@Table(name = "TEST", schema = "L2S$OWNER", catalog = "")
public class MyLibraryTestEntity extends ActionForm {

    @Access(AccessType.FIELD)
    @OneToMany(mappedBy = "sabriEntity", cascade = CascadeType.PERSIST)
    private final List<MyLibrarySabriEntity> entities = new ArrayList<MyLibrarySabriEntity>(25);

    public void addEntitysabri(MyLibrarySabriEntity entity) {
        getEntities().add(entity);
        entity.setSabriEntity(this);
    }

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "IdGenerated")
    @GenericGenerator(name = "IdGenerated", strategy = "dao.Identifier")
    @Column(name = "ID_TRANSITION_MATRIX", nullable = false, insertable = false, updatable = false, length = 10)
    private String ID_TRANSITION_MATRIX;


    @Basic
    @Column(name = "REFERENCE", nullable = true, insertable = true, updatable = true, precision = 0)
    private Integer reference;}

Answer 1

假设您使用的是JPA 2.0 +

完全删除此映射：

@Id
@Column(name = "TRANSITION_MATRIX_ID", nullable = false, 
       insertable = true, updatable = true, length = 100)
private String TRANSITION_MATRIX_ID;

并将@Id直接放在ManyToOne上，删除可插入和可更新的属性。

@Access(AccessType.FIELD)
@Id
@ManyToOne
@JoinColumn(name = "TRANSITION_MATRIX_ID", referencedColumnName = "ID_TRANSITION_MATRIX")
private MyLibraryTestEntity sabriEntity;

相应地更新您的ID类。之前对TRANSITION_MATRIX_ID的任何引用都应替换为对sabriEntity的引用。您还混淆了@EmbeddedId和@IdClass：只有前者包含列定义，而您使用的是后一种方法。

public class MyLibrarySabriEntityPK implements Serializable {

    private String sabriEntity;
    private String RATING_ID_ROW;
    private String RATING_ID_COL;
}

请参阅：

https://en.wikibooks.org/wiki/Java_Persistence/Identity_and_Sequencing#JPA_2.0

Answer 2

感谢Alan Hay，我发现了问题，我将我的IDclass的属性TRANSITION_MATRIX_ID更改为sabriEntity，我删除了这个类的所有注释！

子实体

@Data
@Entity
@IdClass(MyLibrarySabriEntityPK.class)
@Table(name = "SABRI", schema = "L2S$OWNER", catalog = "")
public class MyLibrarySabriEntity extends ActionForm {

@Access(AccessType.FIELD)
@ManyToOne
@Id
@JoinColumn(name = "TRANSITION_MATRIX_ID", referencedColumnName = "ID_TRANSITION_MATRIX")
private MyLibraryTestEntity sabriEntity;


@Id
private String RATING_ID_ROW;

@Id
private String RATING_ID_COL;


@Basic
@Column(name = "TRANSITION_PROBABILITY", nullable = true, insertable = true, updatable = true, precision = 20)
private Double TRANSITION_PROBABILITY;

家长实体

@Data
@Entity
@Table(name = "TEST", schema = "L2S$OWNER", catalog = "")
public class MyLibraryTestEntity extends ActionForm {

@Access(AccessType.FIELD)
@OneToMany(mappedBy = "sabriEntity", cascade = CascadeType.PERSIST)
private final List<MyLibrarySabriEntity> entities = new ArrayList<MyLibrarySabriEntity>(25);

public void addEntitysabri(MyLibrarySabriEntity entity) {
    getEntities().add(entity);
    entity.setSabriEntity(this);
}

@Id
@GeneratedValue(strategy = GenerationType.AUTO, generator = "IdGenerated")
@GenericGenerator(name = "IdGenerated", strategy = "dao.Identifier")
@Column(name = "ID_TRANSITION_MATRIX", nullable = false, insertable = false, updatable = false, length = 10)
private String ID_TRANSITION_MATRIX;


@Basic
@Column(name = "REFERENCE", nullable = true, insertable = true, updatable = true, precision = 0)
private Integer reference;

PK等级

@Data
public class MyLibrarySabriEntityPK implements Serializable {
private MyLibraryTestEntity sabriEntity;
private String RATING_ID_ROW;
private String RATING_ID_COL;

public MyLibrarySabriEntityPK() {
}

public MyLibrarySabriEntityPK(MyLibraryTestEntity sabriEntity,String RATING_ID_COL,String RATING_ID_ROW ){
this.sabriEntity=sabriEntity;
this.RATING_ID_COL = RATING_ID_COL;
this.RATING_ID_ROW= RATING_ID_ROW;

}

使用相同的生成ID保留2个表

2 个答案: