Question

我试图在数据库中进行简单选择，但出现此错误：

Failure while trying to resolve exception [org.springframework.http.converter.HttpMessageNotWritableException]

要创建多对多关系，我遵循以下响应：https://stackoverflow.com/a/32920550/10025420并且我的表还可以。

CommunityLine.java

@Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "community_line_seq")
    @JsonBackReference
    @SequenceGenerator(name = "community_line_seq", sequenceName = "community_line_seq", allocationSize = 1)
    private Long id;

    @Column(name = "apiKey", updatable = true, nullable = false, unique = true, columnDefinition = "BINARY(16)")
    @org.hibernate.annotations.Type(type = "org.hibernate.type.UUIDBinaryType")
    private UUID apiKey;

    @Column(name = "name")
    private String name;

    @ManyToOne
    @JoinColumn(name = "cv_id")
    private Instrument instrument;

    @ManyToOne
    @JoinColumn(name = "param_id")
    private Param param;

    @ManyToOne
    @JoinColumn(name = "sample_type_id")
    private SampleType sampleType;

    @ManyToOne
    @JoinColumn(name = "context_source_id")
    private ContextSource contextSource;

    @Column(name = "value")
    private float value;

    @JsonManagedReference
    @OneToMany(mappedBy = "node", fetch = FetchType.EAGER)
    private List<CommunityLineNodeRelation> nodes;
// Getters and constructors

Node.java

@Entity
@Table(name="node")
public class Node {

    @Id
    @Column(name = "ID")
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "node_seq")
    @SequenceGenerator(name = "node_seq", sequenceName = "node_seq", allocationSize = 1)
    @JsonBackReference
    private Long id;

    @Column(name = "apiKey", updatable = false, nullable = false, unique=true, columnDefinition = "BINARY(16)")
    @org.hibernate.annotations.Type(type="org.hibernate.type.UUIDBinaryType")
    private UUID apiKey;

    @Column
    private String country;

    public String getCountry() {
        return country;
    }

    public void setCountry(String country) {
        this.country = country;
    }

    public UUID getApiKey() {
        return apiKey;
    }


    public void setApiKey(UUID apiKey) {
        this.apiKey = apiKey;
    }

    @Column(name = "NAME", length = 50, unique = true)
    @NotNull
    private String name;


    @OneToMany(mappedBy="node")
    private List<DataSource> dataSource;

    @OneToMany(mappedBy="node", cascade=CascadeType.ALL)
    private List<User> users;

    public List<User> getUsers() {
        return users;
    }

    @JsonBackReference
    @OneToMany(mappedBy = "communityLine", cascade = CascadeType.ALL)
    private List<CommunityLineNodeRelation> communityLines;

//getters and setters

CommunityLineNodeRelation.java

@Entity
@Table(name = "community_line_node")
public class CommunityLineNodeRelation implements Serializable {
    @Id
    @ManyToOne (fetch = FetchType.EAGER)
    @JsonManagedReference
    @JoinColumn(name = "community_line_id", referencedColumnName = "id")
    private CommunityLine communityLine;

    @Id
    @ManyToOne (fetch = FetchType.EAGER)
    @JsonManagedReference
    @JoinColumn(name = "node_id", referencedColumnName = "id")
    private Node node;

    @Column (name = "is_active")
    private boolean isActive;

CommunityLineNodeId.java

public class CommunityLineNodeId implements Serializable {

    private Long communityLine;
    private Long node;

回购

@Repository
public interface CommunityLineRepository extends CrudRepository<CommunityLine, Long> {

    public List<CommunityLine> findAll();

}

哪个类需要存储库？我已经尝试使用CommunityLine和关系类。

为什么会出现此错误？

编辑：我已经设法避免了该错误，但是现在JSON缺少node属性，因此选择有效，因为我可以在服务器控制台中打印该节点。

在Spring Boot中无法通过额外的属性关系来大量获取数据

0 个答案: