在数据库中插入实体+外键的正确方法（使用：Mysql,JPA）

Question

我有一个问题，关于正确的方法，使插入我的实体+外键在数据库(Mysql，jpa)。在问我的问题之前，我尝试了我找到的解决方法，我花了几天时间自己解决这个问题。我好像有什么不对劲:

我需要做的是:
1。站点表中已经存在数据（信息）
2。我必须在序列表中插入新序列
3。我必须在序列表中插入新序列，并关联FK站点的（站点已存在，不需要再次插入）

主要问题是:
1。当我插入序列表时，我有正确插入站点FK的序列
。但是！问题是我在数据库中也有站点插入，所以在插入序列数据时站点的冗余插入

我只需要插入一个有关联的序列（站点的FK）。

下面是我的控制台上发生的情况：

Hibernate: insert into Sites (Current_Sequence, Date_Insert, Date_Update, Enable, Nom_Site, User_Update, applicationService_Id_SA, partner_Id_Partner, Id_Site) values (?, ?, ?, ?, ?, ?, ?, ?, ?)


Hibernate: insert into Sequences (Date_Insert_timestamp, Date_Update_timestamp, Nbr_PDF, Nbr_ZPL, Size_PDF, Size_ZPL, TimestampMinute, User_Update, site_Id_Site, status, Id_Sequence) values (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)


Hibernate: insert into Sites_Sequences (Site_Id_Site, sequences_Id_Sequence) values (?, ?)

Jpa存储库：我首先按名称获取站点对象

    @Override
        public Site findSiteByName(String Name_Site) {
            List sites = entityManager.createQuery("SELECT s FROM Site s").getResultList();
            for (Site item : sites) {
                if (item.getNom_Site().equals(Name_Site)) {
                    return item;
                }
            }

            return null;
        }

Jpa方法在数据库中插入序列

    @Override
        public Sequence saveSequence(Sequence sequence) {

            return super.save(sequence);
        }

我的Rest控制器:

    @GET
    @Path("/addSeq")
    @Produces(MediaType.APPLICATION_JSON)
    @Consumes(MediaType.APPLICATION_JSON)
    public Sequence addSeq(@Context final UriInfo uriInfo) {
        Site site = siteService.findSiteByName("Name_Site");
        Sequence sequence = sequenceFactory.CreateSequence(new Date(), new Date(), "", "", "", 4, 6, 10, 12);
        sequence.setSite(site);
        site.getSequences().add(sequence);
        sequenceservice.saveSequence(sequence);
        return null;
    }

我的主要类是:

    @Entity
    @Table(name = "Sequences")
    public class Sequence {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "Id_Sequence", unique = true, nullable = false)
    private long Id_Sequence;
    private Date ....;
    private Date ....;
    private String ....;
    private String ....;
    private String ....;
    private int ....;
    private int ....;
    private double ....;
    private double ....;
    @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    private Set traceSequences = new HashSet();
    @ManyToOne(cascade = CascadeType.ALL)
    private Site site;

    public constructor...

    public set..
    public get..

 @Entity
    @Table(name = "Sites")
    public class Site  {


    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "Id_Site", unique = true, nullable = false)
    private long Id_Site;
    private String ...;
    private boolean ...;
    private long ...; 
    private Date ...;
    private Date ...;
    private String ...;

    @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    private Set sequences = new HashSet();

    @ManyToOne
    private ... ...;

    @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    private Set ... = new HashSet();

    @ManyToOne
    private ... ...;

    public constructor...

    public set..
    public get..

Answer 1

可以尝试删除所有关系中的cascade=cascadeType.all吗？我认为主要问题是cascadetype.persist被级联到子元素。您可以在这里看到JPA with JTA：Persist entity and merge cascaded child entities

我建议的另一件事是将您的表数从3减少到2。因为您使用的是OneTomany-关系，它不需要像现在这样的映射表。为此，在ManyToOne站点上的OneToMany-annotation和一个附加的JoinColumn-annotation中使用了mappedBy-clause。

@Entity
@Table(name = "Sites")
public class Site {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "Id_Site", unique = true, nullable = false)
    private long Id_Site;
    private String ...;
    private boolean ...;
    private long ...; 
    private Date ...;
    private Date ...;
    private String ...;

    @OneToMany(mappedBy="site", fetch = FetchType.EAGER)
    private Set<Sequence> sequences = new HashSet<>();


@Entity
@Table(name = "Sequences")
public class Sequence {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "Id_Sequence", unique = true, nullable = false)
    private long Id_Sequence;
    private Date ....;
    private Date ....;
    private String ....;
    private String ....;
    private String ....;
    private int ....;
    private int ....;
    private double ....;
    private double ....;

    @OneToMany(fetch = FetchType.EAGER)
    private Set<Sequence> traceSequences = new HashSet<>();

    @ManyToOne(cascade = CascadeType.MERGE)
    @JoinColumn(name = "site_Id_Site", nullable = false)
    private Site site;

编辑我尝试使用：

public interface SequenceRepository extends JpaRepository<Sequence, Long> {

}

而且

Site site = siteRepository.findOne(1L);

Sequence seq = new Sequence();
seq.setSite(site);
site.getSequences().add(seq);

sequenceRepository.save(seq);

带来

Hibernate: select site0_.id_site...
Hibernate: insert into sequences (site_id_site) values (?)

在类路径中，我有spring-data-rest-webmvc-2.5.6.release.jar。