已完成500 INTERNAL\u SERVER\u ERROR SpringBoot Angular:OneToMany问题

Question

evrey我尝试加载列表RDV时遇到此问题：

**类型定义错误：[简单类型，类org.hibernate.proxy.pojo.bytebuddy.ByteBuddyInterceptor]；嵌套异常为com。fasterxml。杰克逊。数据绑定。exc.InvalidDefinitionException：找不到类org的序列化程序。冬眠代理波乔。拜特巴迪。ByteBuddyInterceptor，未发现创建BeanSerializer的属性（为了避免异常，请禁用SerializationFeature.FAIL\u ON\u EMPTY\u bean）（通过引用链：java.util.ArrayList[0]-

o、 s.web。servlet。DispatcherServlet：已完成500 INTERNAL\u SERVER\u错误**

但是，当我删除OneTo许多它的工作！

这是我的客户端类：

public class Client implements Serializable {

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long        id;
private String      name;
private String      prenom;

@JsonIgnoreProperties({ "hibernateLazyInitializer", "handler" })
@OneToMany(mappedBy = "client", fetch = FetchType.LAZY)
private List<Rv>    lrdvs;

@JsonIgnore
public List<Rv> getLrdvs() {
    return lrdvs;
}

@JsonIgnore
public void setLrdvs(List<Rv> lrdvs) {
    this.lrdvs = lrdvs;
}

}

这是Medecin等级：

 @Entity
@Data
@AllArgsConstructor
@NoArgsConstructor
@Table(name = "medecins")
public class Medecin implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private long        id;
private String      nom;
private String      prenom;
private String      specialite;
@JsonIgnoreProperties({ "hibernateLazyInitializer", "handler" })
@OneToMany(mappedBy = "medecin", fetch = FetchType.LAZY)
private List<Rv>    lrdvs;

@JsonIgnore
public List<Rv> getLrdvs() {
    return lrdvs;
}

@JsonIgnore
public void setLrdvs(List<Rv> lrdvs) {
    this.lrdvs = lrdvs;
}

}

这是我的RDV课程

@Entity
@Data
@AllArgsConstructor
public class Rv {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private long        id;
@DateTimeFormat(pattern = "yyyy-MM-dd")
private LocalDate   jour;

@JoinColumn(name = "ID_Client")
@ManyToOne(fetch = FetchType.LAZY, optional = false)
private Client      client;

@JoinColumn(name = "ID_Medecin")
@ManyToOne(fetch = FetchType.LAZY, optional = false)
private Medecin     medecin;

public Rv() {
    client = new Client();
    medecin = new Medecin();
}

}

任何人都知道如何解决它。thx：）

Answer 1

将此行添加到您的应用程序属性

spring.jackson.serialization.fail-on-empty-beans=false