如何防止我的多对多关系无限递归或JSON写入失败？

Question

我试图使双向多关系与 Hibernate 和 Maven 一起工作，但它总是抛出 org.hibernate.LazyInitializationException: 无法初始化代理 - 无会话或 Stackoverflow 错误。我已经尝试过@JSONIdentityInfo，这导致Maven自动添加一个“PK”变量，这主要破坏了我的前端，@JSONManagedReference和@JSONBackreference，这导致接收反向引用的实体被完全忽略，@JSONIgnore，与反向引用相同的问题，@JsonView，它看起来没有任何改变，@JsonSerialize带有自定义序列化器，它在实现时也会导致前端出现问题。我还尝试将查询直接写入存储库，但它导致了同样的问题。我真的已经力竭了，不知道还能做什么。

节点

import com.fasterxml.jackson.annotation.JsonBackReference;
import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
import com.fasterxml.jackson.annotation.JsonInclude;
import com.fasterxml.jackson.annotation.JsonView;
import com.fasterxml.jackson.databind.annotation.JsonSerialize;
import de.sadgmbh.spring.angular.backenddemo.model.AbstractAuditingEntity;
import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;

import javax.persistence.*;
import javax.validation.constraints.NotNull;
import java.util.HashSet;
import java.util.Set;

@Getter
@Setter
@NoArgsConstructor
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonInclude(JsonInclude.Include.NON_NULL)

@Entity
@Table(name = "knotenpunkte")
//@JsonSerialize(using = CustomKnotenpunktSerializer)
public class Knotenpunkt extends AbstractAuditingEntity<Long> {
    @NotNull
    @Column(length = 50, unique = true, nullable = false)
    @JsonView(Views.Public.class)
    private int knotennr;
    @NotNull
    @Column(length = 50, unique = true, nullable = false)
    @JsonView(Views.Public.class)
    private String strasse;
    @NotNull
    @JsonView(Views.Public.class)
    private boolean codierung;
    @NotNull
    @JsonView(Views.Public.class)
    private boolean bake;
    @ManyToMany(fetch = FetchType.EAGER)
    @JoinTable(
            name = "knotenpunkt_linie",
            joinColumns = {@JoinColumn(name = "knotenpunkt_id", referencedColumnName = "id")},
            inverseJoinColumns = {@JoinColumn(name = "linie_id", referencedColumnName = "id")})
    @JsonView(Views.Internal.class)
    Set<Linie> linienSet = new HashSet<>();
}

线

import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
import com.fasterxml.jackson.annotation.JsonInclude;
import com.fasterxml.jackson.annotation.JsonView;
import com.fasterxml.jackson.databind.annotation.JsonSerialize;
import de.sadgmbh.spring.angular.backenddemo.model.AbstractAuditingEntity;
import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;

import javax.persistence.*;
import javax.validation.constraints.NotNull;
import java.util.HashSet;
import java.util.Set;

@Getter
@Setter
@NoArgsConstructor
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonInclude(JsonInclude.Include.NON_NULL)
@Entity
@Table(name = "linien")
//@JsonSerialize(using = CustomLineSerializer.class)
public class Linie extends AbstractAuditingEntity<Long> {
    @NotNull
    @Column(length = 50, unique = true, nullable = false)
    @JsonView(Views.Public.class)
    private String linienNR;
    @ManyToMany( mappedBy = "linienSet")
    @JsonView(Views.Internal.class)
    private Set<Knotenpunkt> knotenpunktSet = new HashSet<>();
}

Maven 生成的接口

export interface Knotenpunkt extends AbstractAuditingEntity<number> {
    id: number;
    knotennr: number;
    strasse: string;
    codierung: boolean;
    bake: boolean;
    linienSet: Linie[];
}

export interface Linie extends AbstractAuditingEntity<number> {
    id: number;
    linienNR: string;
    knotenpunktSet: Knotenpunkt[];
}

export interface Views {
}

export interface Internal extends Public {
}

export interface Public {
}

export interface JsonSerializer<T> extends JsonFormatVisitable {
    unwrappingSerializer: boolean;
    delegatee: JsonSerializer<any>;
}

Answer 1

最终的解决方案是在控制器中添加

@Transactional

。希望这可以帮助某人。

Answer 2

就我而言，我正在做： @JsonManagedRefence 和 @JsonBackReference，这是错误的，因为这些仅用于父子关系。

对于多对多使用： @JsonIdentityInfo（生成器= ObjectIdGenerators.PropertyGenerator.class，属性=“id”） 希望这对某人有帮助。