带有额外列的 JPA 2.0 多对多 - 更新集合

Posted 2023-04-12

【中文标题】带有额外列的 JPA 2.0 多对多 - 更新集合

【英文标题】：JPA 2.0 many-to-many with extra column - Update collection

【发布时间】：2020-05-05 10:12:35

【问题描述】：

我正在使用following example

@Entity
public class Employer 

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;

    @OneToMany(mappedBy = "employer")
    private List<EmployerDeliveryAgent> deliveryAgentAssoc;

    // other properties and getters and setters


@Entity
public class DeliveryAgent 

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;

    @OneToMany(mappedBy = "deliveryAgent")
    private List<EmployerDeliveryAgent> employerAssoc;

    // other properties and getters and setters

关联类

@Entity
@Table(name = "employer_delivery_agent")
@IdClass(EmployerDeliveryAgentId.class)
public class EmployerDeliveryAgent 

    @Id
    @ManyToOne
    @JoinColumn(name = "employer_id", referencedColumnName = "id")
    private Employer employer;

    @Id
    @ManyToOne
    @JoinColumn(name = "delivery_agent_id", referencedColumnName = "id")
    private DeliveryAgent deliveryAgent;

    @Column(name = "is_project_lead")
    private boolean isProjectLead;

公会PK类：

public class EmployerDeliveryAgentId implements Serializable 

    private int employer;
    private int deliveryAgent;

    // getters/setters and most importantly equals() and hashCode()

如何更新List<EmployerDeliveryAgent> deliveryAgentAssoc;？

如果我获得 Employer 实体并执行简单的 setDeliveryAgentAssoc() 并将其设置为新列表并保存 Employer 实体，我最终会在我的数据库中得到旧列表和新列表。

我也尝试过以下代码，但由于某种原因它再次没有删除旧集合：

employer.getDeliveryAgentAssoc().forEach(employerDeliveryAgentRepository::delete);
employer.setDeliveryAgent(newCollection);
employerRepository.save(employer);

我想用新集合替换现有集合的所有内容。我该怎么做？

【参考方案1】：

您缺少Owning 实体的概念。 Employer 和DeliveryAgent 中的mappedBy 注释将关系的拥有实体定义为EmployerDeliveryAgent 实体。由于您自己定义了所有实体和存储库，因此您也必须自己管理它们。

在非所有者实体中设置关系属性不会做任何持久性。因此，这些属性仅供查询。设置 deliveryAgentAssoc 或 employerAssoc 对 JPA 没有任何作用。

此外，ManyToMany 的 Embedded 模式更新，通常效果更好。

最后，使用Set，除非你真的认为你有理由为OneToMany 关系订购List。在一个实体中有多个 List 关系会导致 JPA 问题。此外，不要在基于 JPA 的包中使用 Java 原始类型，因为这会导致问题和混乱。

在编写代码时打印出 SQL 语句，以便您了解发生了什么。这很重要，因为 JPA 会在您不查看时执行大量惰性提取，如果您不了解正在发生的事情并对其进行管理，您最终会遇到您不了解的错误和问题。向任何 JPA 编码人员询问可怕的 LazyInitializationException。

所以，作为建议的结果：

@Entity
public class Employer 

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @OneToMany(mappedBy="employer")
    private Set<EmployerDeliveryAgent> deliveryAgentAssoc;

@Entity
public class DeliveryAgent 

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @OneToMany(mappedBy="deliveryAgent")
    private Set<EmployerDeliveryAgent> employerAssoc;

@Entity
@ToString
public class EmployerDeliveryAgent 

    @EmbeddedId
    private EmployerDeliveryAgentId id = new EmployerDeliveryAgentId();

    @ManyToOne
    @MapsId("employerId")
    private Employer employer;

    @ManyToOne
    @MapsId("deliveryAgentId")
    private DeliveryAgent deliveryAgent;

    @Column(name = "is_project_lead")
    private Boolean isProjectLead;

还有你的 ID 等级：

@Embeddable
public class EmployerDeliveryAgentId implements Serializable 
    private static final long serialVersionUID = 1L;
    private Long employerId;
    private Long deliveryAgentId;

并使用它：

    Employer emp1 = new Employer();
    employerRepo.save(emp1);
    DeliveryAgent da1 = new DeliveryAgent();
    deliveryAgentRepo.save(da1);
    EmployerDeliveryAgent eda1 = new EmployerDeliveryAgent();
    eda1.setEmployer(emp1);
    eda1.setDeliveryAgent(da1);
    eda1.setProjectLead(false);
    employerDeliveryAgentRepo.save(eda1);
    DeliveryAgent da2 = new DeliveryAgent();
    deliveryAgentRepo.save(da2);
    EmployerDeliveryAgent eda2 = new EmployerDeliveryAgent();
    eda2.setEmployer(emp1);
    eda2.setDeliveryAgent(da2);
    eda2.setProjectLead(true);
    employerDeliveryAgentRepo.save(eda2);

    employerDeliveryAgentRepo.findAll().forEach(System.out::println);

    EmployerDeliveryAgent edaex = new EmployerDeliveryAgent();
    edaex.setEmployer(emp1);
    employerDeliveryAgentRepo.findAll(Example.of(edaex)).forEach(System.out::println);

    employerDeliveryAgentRepo.deleteAll( employerDeliveryAgentRepo.findAll(Example.of(edaex)));

导致以下日志输出：

Hibernate: drop table delivery_agent if exists
Hibernate: drop table employer if exists
Hibernate: drop table employer_delivery_agent if exists
Hibernate: create table delivery_agent (id bigint generated by default as identity, primary key (id))
Hibernate: create table employer (id bigint generated by default as identity, primary key (id))
Hibernate: create table employer_delivery_agent (is_project_lead boolean, delivery_agent_id bigint not null, employer_id bigint not null, primary key (delivery_agent_id, employer_id))
Hibernate: alter table employer_delivery_agent add constraint FKqfdjch3412029revbsh103okx foreign key (delivery_agent_id) references delivery_agent
Hibernate: alter table employer_delivery_agent add constraint FKc3djdeycywdtbpn4muakrhhtq foreign key (employer_id) references employer
2020-05-05 10:56:36.200  INFO 7588 --- [           main] o.h.e.t.j.p.i.JtaPlatformInitiator       : HHH000490: Using JtaPlatform implementation: [org.hibernate.engine.transaction.jta.platform.internal.NoJtaPlatform]
2020-05-05 10:56:36.204  INFO 7588 --- [           main] j.LocalContainerEntityManagerFactoryBean : Initialized JPA EntityManagerFactory for persistence unit 'default'
2020-05-05 10:56:36.403  INFO 7588 --- [           main] com.example.demo.DemoApplication         : Started DemoApplication in 1.455 seconds (JVM running for 1.821)
Hibernate: insert into employer (id) values (null)
Hibernate: insert into delivery_agent (id) values (null)
Hibernate: select employerde0_.delivery_agent_id as delivery2_2_0_, employerde0_.employer_id as employer3_2_0_, employerde0_.is_project_lead as is_proje1_2_0_ from employer_delivery_agent employerde0_ where employerde0_.delivery_agent_id=? and employerde0_.employer_id=?
Hibernate: select deliveryag0_.id as id1_0_0_ from delivery_agent deliveryag0_ where deliveryag0_.id=?
Hibernate: select employer0_.id as id1_1_0_ from employer employer0_ where employer0_.id=?
Hibernate: insert into employer_delivery_agent (is_project_lead, delivery_agent_id, employer_id) values (?, ?, ?)
Hibernate: insert into delivery_agent (id) values (null)
Hibernate: select employerde0_.delivery_agent_id as delivery2_2_0_, employerde0_.employer_id as employer3_2_0_, employerde0_.is_project_lead as is_proje1_2_0_ from employer_delivery_agent employerde0_ where employerde0_.delivery_agent_id=? and employerde0_.employer_id=?
Hibernate: select deliveryag0_.id as id1_0_0_ from delivery_agent deliveryag0_ where deliveryag0_.id=?
Hibernate: select employer0_.id as id1_1_0_ from employer employer0_ where employer0_.id=?
Hibernate: insert into employer_delivery_agent (is_project_lead, delivery_agent_id, employer_id) values (?, ?, ?)
Hibernate: select employerde0_.delivery_agent_id as delivery2_2_, employerde0_.employer_id as employer3_2_, employerde0_.is_project_lead as is_proje1_2_ from employer_delivery_agent employerde0_
Hibernate: select deliveryag0_.id as id1_0_0_ from delivery_agent deliveryag0_ where deliveryag0_.id=?
Hibernate: select employer0_.id as id1_1_0_ from employer employer0_ where employer0_.id=?
Hibernate: select deliveryag0_.id as id1_0_0_ from delivery_agent deliveryag0_ where deliveryag0_.id=?
EmployerDeliveryAgent(id=com.example.demo.EmployerDeliveryAgentId@3e1, employer=com.example.demo.Employer@52ae997b, deliveryAgent=com.example.demo.DeliveryAgent@32f32623, isProjectLead=false)
EmployerDeliveryAgent(id=com.example.demo.EmployerDeliveryAgentId@400, employer=com.example.demo.Employer@52ae997b, deliveryAgent=com.example.demo.DeliveryAgent@7e15f4d4, isProjectLead=true)
Hibernate: select employerde0_.delivery_agent_id as delivery2_2_, employerde0_.employer_id as employer3_2_, employerde0_.is_project_lead as is_proje1_2_ from employer_delivery_agent employerde0_ inner join employer employer1_ on employerde0_.employer_id=employer1_.id where employer1_.id=1
Hibernate: select deliveryag0_.id as id1_0_0_ from delivery_agent deliveryag0_ where deliveryag0_.id=?
Hibernate: select employer0_.id as id1_1_0_ from employer employer0_ where employer0_.id=?
Hibernate: select deliveryag0_.id as id1_0_0_ from delivery_agent deliveryag0_ where deliveryag0_.id=?
EmployerDeliveryAgent(id=com.example.demo.EmployerDeliveryAgentId@3e1, employer=com.example.demo.Employer@62b57479, deliveryAgent=com.example.demo.DeliveryAgent@1903b5d, isProjectLead=false)
EmployerDeliveryAgent(id=com.example.demo.EmployerDeliveryAgentId@400, employer=com.example.demo.Employer@62b57479, deliveryAgent=com.example.demo.DeliveryAgent@5a90265a, isProjectLead=true)
Hibernate: select employerde0_.delivery_agent_id as delivery2_2_, employerde0_.employer_id as employer3_2_, employerde0_.is_project_lead as is_proje1_2_ from employer_delivery_agent employerde0_ inner join employer employer1_ on employerde0_.employer_id=employer1_.id where employer1_.id=1
Hibernate: select deliveryag0_.id as id1_0_0_ from delivery_agent deliveryag0_ where deliveryag0_.id=?
Hibernate: select employer0_.id as id1_1_0_ from employer employer0_ where employer0_.id=?
Hibernate: select deliveryag0_.id as id1_0_0_ from delivery_agent deliveryag0_ where deliveryag0_.id=?
Hibernate: select employerde0_.delivery_agent_id as delivery2_2_0_, employerde0_.employer_id as employer3_2_0_, employerde0_.is_project_lead as is_proje1_2_0_, deliveryag1_.id as id1_0_1_, employer2_.id as id1_1_2_ from employer_delivery_agent employerde0_ inner join delivery_agent deliveryag1_ on employerde0_.delivery_agent_id=deliveryag1_.id inner join employer employer2_ on employerde0_.employer_id=employer2_.id where employerde0_.delivery_agent_id=? and employerde0_.employer_id=?
Hibernate: select employerde0_.delivery_agent_id as delivery2_2_0_, employerde0_.employer_id as employer3_2_0_, employerde0_.is_project_lead as is_proje1_2_0_, deliveryag1_.id as id1_0_1_, employer2_.id as id1_1_2_ from employer_delivery_agent employerde0_ inner join delivery_agent deliveryag1_ on employerde0_.delivery_agent_id=deliveryag1_.id inner join employer employer2_ on employerde0_.employer_id=employer2_.id where employerde0_.delivery_agent_id=? and employerde0_.employer_id=?
Hibernate: delete from employer_delivery_agent where delivery_agent_id=? and employer_id=?
Hibernate: delete from employer_delivery_agent where delivery_agent_id=? and employer_id=?

【参考方案2】：

Employer 或 DeliveryAgent 应负责更新 EmployerDeliveryAgent 集合。

在 Employer 类中添加以下方法。

public void addEmployerAssoc(EmployerDeliveryAgent item) 
    item.setEmployer(this);
    deliveryAgentAssoc.add(item);


public void removeEmployerAssoc(EmployerDeliveryAgent item) 
    item.setEmployer(null);
    deliveryAgentAssoc.remove(item); // must implement equals/hashcode

【讨论】：

这并没有回答如何用新集合替换现有集合的所有内容的问题。如何删除现有的集合？为每个元素调用 removeEmployerAssoc？ （导致 ConcurrentModificationException）。只需调用 clear 上的集合？ （什么都不做）

即使删除一个元素也不适用于您的示例。 HibernateException：复合标识符的任何部分都不能为空