Question

此代码失败我在数据库中有一个现存的版本化值，该值是我使用entity manager.find获取的，并在实体上进行了设置。

VersionEntity version = entityManager.find(VersionEntity.class, 1);
entities.forEach(__ -> __.setVersion(version));
entities.forEach(entityManager::persist);
entityManager.flush();

我正在专门查找并分配版本1，所以为什么它甚至尝试使用版本2。我什至尝试了merge，但没有用。我的属性显然是错误的/不正确的。

刷新失败

ocal异常堆栈：异常[EclipseLink-4002]（Eclipse Persistence Services-2.7.1.qualifier）：org.eclipse.persistence.exceptions.DatabaseException 内部异常：org.postgresql.util.PSQLException：错误：在表“ t_quality_measurement”上插入或更新违反了外键约束“ fk_t_quality_measurement_version” 详细信息：表（t_version）中不存在键（版本）=（2）。错误代码：0 呼叫：插入T_QUALITY_MEASUREMENT（品牌，日期，KPICODE，KPICOMPAREVW，KPICOMPAREWT，KPIVALUE，组织单位，相位，工厂，系列，信号颜色，技术，温度，视图，视图）值（？，？，？，？，？，？，？，？？，？，？，？，？，？，？）绑定=> [绑定14个参数] 查询：

我不确定为什么我特意查找1时为什么要使用2。

@Entity
@Table(name = "T_QUALITY_MEASUREMENT")
public class QualityMeasurementEntity extends Versioned implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(columnDefinition = "serial")
    private Integer id;

    @Column(length = 100)
    private String series;

版本看起来像这样

@MappedSuperclass
public abstract class Versioned {
    @ManyToOne
    @JoinColumn(name = "version", insertable = false, updatable = false)
    protected VersionEntity version = null;

    public VersionEntity getVersion() { return version; }

    public void setVersion(VersionEntity version) { this.version = version; }
}

和版本实体。

@Entity
@Table(name = "T_VERSION", indexes = {
        @Index(name = "pk_version", columnList = "id", unique = true)
})
@NamedQuery(name = "Version.findById", query = "SELECT v FROM VersionEntity v WHERE v.id = :id")
public class VersionEntity {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(columnDefinition = "serial")
    private Integer id;

    @Column(name = "CREATED")
    private LocalDateTime created;

    @Column
    private Boolean active = false;

    @PrePersist
    protected void onCreate() {
        created = LocalDateTime.now();
    }

    public Integer getId() {
        return id;
    }
    public void setId(Integer id) {
        this.id = id;
    }

    public LocalDateTime getCreated() { return created; }
    public void setCreated(LocalDateTime created) { this.created = created; }

    public Boolean getActive() { return active; }
    public void setActive(Boolean active) { this.active = active; }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        VersionEntity version = (VersionEntity) o;
        return id.equals(version.id);
    }

    @Override public int hashCode() { return id.hashCode(); }
}

Answer 1

There is no need to prevent the insert. Change @JoinColumn

 @JoinColumn(name = "version")

JPA持久性级联孩子时插入错误的ID值

1 个答案: