Question

在一个带有hibernate和jpa的spring boot 2项目中，

我尝试保存一些对象

@Embeddable
public class EmbedddedSamplesKey implements Serializable {

    private Integer id;
    private int year;
    ...
}

@Entity
@IdClass(EmbedddedSamplesKey.class)
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public class Samplings {

    @Id
    @GeneratedValue
    private Integer id;

    @Id
    private int year;

    @OneToMany(mappedBy = "sampling", cascade = CascadeType.PERSIST, orphanRemoval = true)
    private List<Samples> samples = new ArrayList<>();

    public void addSample(Samples sample) {
      samples.add(sample);
      sample.setSampling(this);
    }

    public void removeSample(Samples sample) {
      samples.remove(sample);
      sample.setSampling(null);
    }

    ...
}

@Entity
public class Samples extends BaseEntity {
    @Id
    @SequenceGenerator(name = "samples_id_seq", sequenceName = "samples_id_seq", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "samples_id_seq")
    private Integer id;

    private String letter;

    //@OneToOne(optional = false, cascade = CascadeType.PERSIST)
    //private TestSamples testSamples;

    @ManyToOne
    private Samplings sampling;
    ...
}

当我保存采样时，我会在样本表中看到

sampling_id sampling_year

不是Feed

这应该是这样做的工作吗？

在样本中，orm生成了这个

ALTER TABLE samples
  ADD CONSTRAINT fkq5if151jgtlcy7yfp55ffvf47 FOREIGN KEY (sampling_id, sampling_year)
  REFERENCES samplings (id, year)
  ON UPDATE NO ACTION
  ON DELETE NO ACTION;

编辑，尝试

@ManyToOne
@JoinColumns({
    @JoinColumn(name = "sampling_id", referencedColumnName = "id"),
    @JoinColumn(name = "sampling_year", referencedColumnName = "year")})
private Samplings sampling;

但结果相同

另外，如果在样本中我想使用像主键那样的采样主+字段字母有没有办法做到这一点？

Answer 1

它可能与尼古拉斯所说的相反

请查看此示例

https://github.com/hibernate/hibernate-orm/blob/ceaeb81e3362ff187004ea3479b2afeeba5aa8a6/documentation/src/test/java/org/hibernate/userguide/mapping/identifier/IdClassGeneratedValueTest.java#L77

更改为cascade.all，应该有帮助...可能会将数据sprind到合并而不是持久化。

可能是春季启动问题？

Answer 2

也许不是最好的答案，但最好等3个月才能得到答案....时间就是金钱

如果不知道需要在序列上做任何特殊操作....

只需使用序列，添加一个瞬态字段，该字段将是序列+年....

Answer 3

粗略地看一眼，我认为你不能将@GeneratedValue与@IdClass一起使用。

Auto Increment not working for Entity class with composite Key

hibernate

3 个答案: