无法创建具有可嵌入ID的新实体

Question

我有一个实体

@Entity
public class EcranChamp {

@EmbeddedId
@AttributeOverrides({ @AttributeOverride(name = "id_ecran", column = @Column(name = 
    "ecran", nullable = false)),
@AttributeOverride(name = "id_champ", column = @Column(name = "champ", nullable = false)) })
EcranChampId id ;

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "id_ecran")
Ecran ecran;

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "id_champ")
Champ champ; 
//getters & setters 

@Embeddable
public class EcranChampId implements Serializable {

 @Column(name = "id_champ", nullable = false) 
  private Long id_champ; 
  @Column(name = "id_ecran", nullable = false)  
  private Long id_ecran; 

EcranChamp ec = new EcranChamp();
ec.setChamp(c);
ec.setEcran(e);

我有这个错误

Answer 1

尝试将Champ和Ecran放入可嵌入id中

@Embeddable
public class EcranChampId implements Serializable {
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "id_champ")
    Champ champ; 

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "id_ecran")
    Ecran ecran;
    //getters & setters 
}


EcranChamp ec = new EcranChamp();
EcranChampId  id = new EcranChampId();
id.setChamp(c);
id.setEcran(e);
ec.setId(id);

并且必须在Ecran和Champ中使用@id@generatedvalue(strategy=generationtype.identity)